Chapter 7 Triangles (Additional Questions)

Two geometric figures are said to be congruent if they have exactly the same size and shape. This means that one figure can be placed exactly on top of the other, covering it completely. They are exact copies of each other.

Let's consider the given options:

(A) Having the same area does not guarantee congruence. For example, a triangle with base 6 and height 4 has an area of $\frac{1}{2} \times 6 \times 4 = 12$, and a triangle with base 8 and height 3 also has an area of $\frac{1}{2} \times 8 \times 3 = 12$, but these triangles are not necessarily congruent.

(B) Having the same perimeter does not guarantee congruence. For instance, a right-angled triangle with sides 3, 4, and 5 has a perimeter of $3+4+5=12$. An isosceles triangle with sides 5, 5, and 2 also has a perimeter of $5+5+2=12$, but these triangles are not congruent.

(D) Having corresponding angles equal defines similar triangles. Similar triangles have the same shape but can be different in size. For congruence, the size must also be the same.

(C) The definition of congruent figures is that they are exact replicas and can be superimposed on each other, meaning they can be made to coincide perfectly. This option directly states the definition of congruence.

Therefore, two triangles are congruent if they are exact copies of each other and can be made to coincide.

The correct option is (C) They are exact copies of each other and can be made to coincide.

There are several criteria used to determine if two triangles are congruent without having to check if all six corresponding parts (three sides and three angles) are equal. The common congruence criteria are:

• SSS (Side-Side-Side): If the three sides of one triangle are respectively equal to the three sides of another triangle, then the triangles are congruent.
• SAS (Side-Angle-Side): If two sides and the included angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the triangles are congruent.
• ASA (Angle-Side-Angle): If two angles and the included side of one triangle are respectively equal to two angles and the included side of another triangle, then the triangles are congruent.
• AAS (Angle-Angle-Side): If two angles and a non-included side of one triangle are respectively equal to two angles and the corresponding non-included side of another triangle, then the triangles are congruent. (Note: This can be derived from ASA using the angle sum property of a triangle).
• RHS (Right-angle-Hypotenuse-Side): If in two right-angled triangles, the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle, then the triangles are congruent.

The condition SSA (Side-Side-Angle), where two sides and a non-included angle are equal, is not a valid congruence criterion. This is because it can lead to two different possible triangles (the ambiguous case), meaning the triangles are not necessarily congruent even if these three parts are equal.

Therefore, SSA is NOT a criterion for the congruence of two triangles.

The correct option is (C) SSA.

In the given problem, we are provided with the following information about $\triangle ABC$ and $\triangle PQR$:

1. Side $AB$ of $\triangle ABC$ is equal to side $PQ$ of $\triangle PQR$. ($AB = PQ$)

2. Angle $\angle B$ of $\triangle ABC$ is equal to angle $\angle Q$ of $\triangle PQR$. ($\angle B = \angle Q$)

3. Side $BC$ of $\triangle ABC$ is equal to side $QR$ of $\triangle PQR$. ($BC = QR$)

We observe that the equal angles $\angle B$ and $\angle Q$ are the angles included between the pairs of equal sides ($AB, BC$ and $PQ, QR$) respectively.

This configuration of two sides and the included angle corresponds to the SAS (Side-Angle-Side) congruence criterion.

The SAS congruence criterion states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent.

Therefore, based on the given conditions, $\triangle ABC$ is congruent to $\triangle PQR$ by the SAS criterion.

$\triangle ABC \cong \triangle PQR \quad \text{(By SAS congruence criterion)}$

Let's briefly look at the other options to confirm why they are not applicable here:

• SSS: Requires three pairs of corresponding sides to be equal. We are only given two pairs of equal sides.
• ASA: Requires two pairs of corresponding angles and the included side to be equal. We are given one pair of equal angles and two pairs of equal sides.
• RHS: Applies specifically to right-angled triangles, requiring the hypotenuse and one side to be equal. We are not told that the triangles are right-angled.

Thus, the only applicable criterion among the given options is SAS.

The correct option is (C) SAS.

We are given that $\triangle ABC \cong \triangle PQR$.

The abbreviation CPCT stands for Corresponding Parts of Congruent Triangles. This principle states that if two triangles are congruent, then their corresponding sides and corresponding angles are equal.

When the congruence of two triangles is stated in the form $\triangle ABC \cong \triangle PQR$, the order of the vertices is crucial. The order indicates the correspondence between the vertices of the two triangles.

In the statement $\triangle ABC \cong \triangle PQR$:

The first vertex of the first triangle ($A$) corresponds to the first vertex of the second triangle ($P$).

The second vertex of the first triangle ($B$) corresponds to the second vertex of the second triangle ($Q$).

The third vertex of the first triangle ($C$) corresponds to the third vertex of the second triangle ($R$).

Based on this vertex correspondence, the corresponding parts are:

• Corresponding angles: $\angle A$ corresponds to $\angle P$, $\angle B$ corresponds to $\angle Q$, and $\angle C$ corresponds to $\angle R$.
• Corresponding sides: Side $AB$ (formed by the first two vertices) corresponds to side $PQ$ (formed by the first two vertices), side $BC$ (formed by the second and third vertices) corresponds to side $QR$ (formed by the second and third vertices), and side $AC$ (formed by the first and third vertices) corresponds to side $PR$ (formed by the first and third vertices).

By the CPCT principle, the corresponding parts are equal. Therefore:

$\angle A = \angle P$

$\angle B = \angle Q$

$\angle C = \angle R$

$AB = PQ$

$BC = QR$

$AC = PR$

The question asks for the angle equal to $\angle A$. From the correspondence established by the congruence statement, $\angle A$ corresponds to $\angle P$. Therefore, by CPCT, $\angle A = \angle P$.

The correct option is (A) $\angle P$.

An isosceles triangle is defined as a triangle that has at least two sides of equal length. The third side is often called the base.

A fundamental property of isosceles triangles is that the angles opposite the equal sides are equal.

Consider an isosceles triangle $\triangle ABC$ where $AB = AC$. The angle opposite side $AB$ is $\angle C$, and the angle opposite side $AC$ is $\angle B$. The property states that if $AB = AC$, then $\angle B = \angle C$. These are often referred to as the base angles of the isosceles triangle.

This property is a direct consequence of the SAS congruence criterion. If we consider the angle bisector of the angle between the equal sides (the apex angle), it divides the isosceles triangle into two congruent right-angled triangles, proving that the base angles must be equal.

Let's examine the options:

• (A) Complementary angles sum up to $90^\circ$. This is not a general property of the base angles of an isosceles triangle.
• (B) Supplementary angles sum up to $180^\circ$. This is the sum of all angles in a triangle, not specifically the relationship between the angles opposite the equal sides.
• (C) Equal: This matches the established property of isosceles triangles.
• (D) Right angles ($90^\circ$): This would only be true if the isosceles triangle is also a right-angled triangle, which is a specific case (a right isosceles triangle), not a general property for all isosceles triangles.

Thus, in an isosceles triangle, the angles opposite the equal sides are always equal.

The correct option is (C) Equal.

We are given a triangle $\triangle ABC$ with the condition $AB = AC$.

A triangle in which two sides are equal is called an isosceles triangle.

In an isosceles triangle, there is a property that states that the angles opposite the equal sides are equal.

In $\triangle ABC$, the side $AB$ is equal to the side $AC$.

The angle opposite to side $AB$ is $\angle C$.

The angle opposite to side $AC$ is $\angle B$.

According to the property of isosceles triangles, since $AB = AC$, the angles opposite these sides must be equal.

Therefore, $\angle C = \angle B$, or equivalently, $\angle B = \angle C$.

Let's look at the given options:

• (A) $\angle B = \angle C$: This matches our conclusion based on the property of isosceles triangles.
• (B) $\angle A = \angle B$: This would imply that the side opposite $\angle A$ ($BC$) is equal to the side opposite $\angle B$ ($AC$), i.e., $BC = AC$. This is not necessarily true for a triangle where only $AB = AC$.
• (C) $\angle A = \angle C$: This would imply that the side opposite $\angle A$ ($BC$) is equal to the side opposite $\angle C$ ($AB$), i.e., $BC = AB$. This is also not necessarily true.
• (D) $\angle A + \angle B = 90^\circ$: This is not a general property of isosceles triangles. It would mean the third angle, $\angle C$, is $180^\circ - 90^\circ = 90^\circ$, implying it's a right-angled triangle, which is a specific type of isosceles triangle, not all of them.

Thus, if $AB = AC$ in $\triangle ABC$, then $\angle B = \angle C$.

The correct option is (A) $\angle B = \angle C$.

In any triangle, there is a direct relationship between the measures of the angles and the lengths of the sides opposite to them.

The property states that in a triangle, the side opposite to the largest angle is the largest side, and similarly, the angle opposite to the largest side is the largest angle.

Conversely, the side opposite to the smallest angle is the smallest side, and the angle opposite to the smallest side is the smallest angle.

Consider a triangle $\triangle ABC$. Let the angles be $\angle A, \angle B, \angle C$ and the sides opposite to these angles be $a$ (side $BC$), $b$ (side $AC$), and $c$ (side $AB$) respectively.

If, for example, $\angle A > \angle B > \angle C$, then it follows that the side opposite $\angle A$ ($a$) is greater than the side opposite $\angle B$ ($b$), which is greater than the side opposite $\angle C$ ($c$). That is, $a > b > c$.

This principle is a fundamental theorem in geometry and is often used in proving inequalities in triangles.

Therefore, the side opposite to the largest angle is the largest side.

Let's check the options:

• (A) Smallest: Incorrect. The side opposite the smallest angle is the smallest side.
• (B) Largest: Correct. This aligns with the property.
• (C) Equal to the smallest: Incorrect. Only in specific cases like an equilateral triangle where all sides and angles are equal.
• (D) Perpendicular: This term relates to angles ($90^\circ$) or lines/segments, not the relative length of a side opposite an angle.

The correct option is (B) Largest.

We are given a triangle $\triangle ABC$ where $\angle B = 90^\circ$. This means $\triangle ABC$ is a right-angled triangle, with the right angle at vertex $B$.

In any triangle, the side opposite the largest angle is the largest side, and the angle opposite the largest side is the largest angle.

In $\triangle ABC$, the sum of the angles is $180^\circ$. So, $\angle A + \angle B + \angle C = 180^\circ$.

Given $\angle B = 90^\circ$, we have $\angle A + 90^\circ + \angle C = 180^\circ$.

This implies $\angle A + \angle C = 180^\circ - 90^\circ = 90^\circ$.

Since $\angle A$ and $\angle C$ are positive angles (angles of a triangle must be greater than $0^\circ$), it follows that $\angle A < 90^\circ$ and $\angle C < 90^\circ$.

Comparing the angles $\angle A$, $\angle B$, and $\angle C$, we see that $\angle B = 90^\circ$ is greater than both $\angle A$ and $\angle C$. Thus, $\angle B$ is the largest angle in $\triangle ABC$.

According to the property relating angles and sides, the side opposite the largest angle is the longest side.

The side opposite to $\angle B$ is the side $AC$.

Therefore, $AC$ is the longest side of $\triangle ABC$. In a right-angled triangle, the side opposite the right angle is also known as the hypotenuse, which is always the longest side.

The correct option is (C) AC.

This question pertains to a fundamental property of triangles known as the Triangle Inequality Theorem.

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Consider a triangle with side lengths $a$, $b$, and $c$. The theorem can be expressed mathematically as:

$a + b > c$

$a + c > b$

$b + c > a$

This property must hold true for any valid triangle. If the sum of two sides were less than or equal to the third side, the three segments could not form a triangle; they would either be too short to meet or would lie flat along a line.

Let's look at the given options in the context of the Triangle Inequality Theorem:

• (A) Less: This contradicts the theorem. The sum of two sides must be greater than the third.
• (B) Equal: This would mean the three points are collinear (lie on a straight line), forming a degenerate triangle, not a standard triangle where the vertices are non-collinear.
• (C) Greater: This aligns perfectly with the Triangle Inequality Theorem.
• (D) Double: This is not a required relationship. The sum might be double, but it's only required to be greater.

Therefore, the sum of any two sides of a triangle is always greater than the third side.

The correct option is (C) Greater.

We are given a triangle $\triangle ABC$ with the condition $AB < BC$.

In any triangle, there is a property that relates the lengths of the sides to the measures of the angles opposite them.

The property states: The angle opposite to the longer side is greater, and the angle opposite to the shorter side is smaller.

Conversely, the side opposite to the larger angle is longer, and the side opposite to the smaller angle is shorter.

In $\triangle ABC$, the side $AB$ is opposite to the angle $\angle C$, and the side $BC$ is opposite to the angle $\angle A$.

We are given that $AB < BC$.

Since $BC$ is longer than $AB$, the angle opposite to $BC$ must be greater than the angle opposite to $AB$.

The angle opposite to $BC$ is $\angle A$.

The angle opposite to $AB$ is $\angle C$.

Therefore, since $BC > AB$, we have $\angle A > \angle C$.

Let's examine the given options:

• (A) $\angle A < \angle C$: This means $\angle C$ is greater than $\angle A$, which would imply $AB > BC$, contradicting the given condition.
• (B) $\angle A > \angle C$: This means $\angle A$ is greater than $\angle C$, which implies the side opposite $\angle A$ ($BC$) is greater than the side opposite $\angle C$ ($AB$). This matches the given condition $BC > AB$ (or $AB < BC$).
• (C) $\angle B < \angle C$: This compares $\angle B$ and $\angle C$. The side opposite $\angle B$ is $AC$. We do not have enough information to compare the length of $AC$ with the length of $AB$, so we cannot conclude a relationship between $\angle B$ and $\angle C$.
• (D) $\angle B > \angle C$: Similar to option (C), we cannot conclude this relationship based only on $AB < BC$.

Thus, if $AB < BC$ in $\triangle ABC$, then $\angle A > \angle C$.

The correct option is (B) $\angle A > \angle C$.

For any set of three positive lengths to form a triangle, they must satisfy the Triangle Inequality Theorem.

The theorem states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.

Let the side lengths be $a$, $b$, and $c$. The following three inequalities must hold:

$a + b > c$

$a + c > b$

$b + c > a$

We will check each option:

(A) 2 cm, 3 cm, 5 cm

Check 1: $2 + 3 > 5 \implies 5 > 5$ (False)

Since the first inequality is false, these lengths cannot form a triangle.

(B) 3 cm, 4 cm, 6 cm

Check 1: $3 + 4 > 6 \implies 7 > 6$ (True)

Check 2: $3 + 6 > 4 \implies 9 > 4$ (True)

Check 3: $4 + 6 > 3 \implies 10 > 3$ (True)

Since all three inequalities are true, these lengths can form a triangle.

(C) 4 cm, 5 cm, 10 cm

Check 1: $4 + 5 > 10 \implies 9 > 10$ (False)

Since the first inequality is false, these lengths cannot form a triangle.

(D) 1 cm, 2 cm, 3 cm

Check 1: $1 + 2 > 3 \implies 3 > 3$ (False)

Since the first inequality is false, these lengths cannot form a triangle.

Only the set of side lengths 3 cm, 4 cm, and 6 cm satisfies the Triangle Inequality Theorem.

The correct option is (B) 3 cm, 4 cm, 6 cm.

A right-angled triangle is a triangle in which one of the angles is a right angle, i.e., equal to $90^\circ$.

In any triangle, there is a relationship between the size of an angle and the length of the side opposite to it. Specifically, the side opposite the largest angle is the longest side.

In a right-angled triangle, let the angles be $\angle A, \angle B, \angle C$. Suppose $\angle B = 90^\circ$.

The sum of the angles in a triangle is $180^\circ$. So, $\angle A + \angle B + \angle C = 180^\circ$.

Substituting $\angle B = 90^\circ$, we get $\angle A + 90^\circ + \angle C = 180^\circ$, which means $\angle A + \angle C = 90^\circ$.

Since $\angle A$ and $\angle C$ are positive angles, both $\angle A < 90^\circ$ and $\angle C < 90^\circ$.

Comparing the angles, we see that $\angle B = 90^\circ$ is greater than both $\angle A$ and $\angle C$. Therefore, $\angle B$ is the largest angle in the right-angled triangle.

The side opposite the right angle ($\angle B$) is called the hypotenuse.

Since the hypotenuse is the side opposite the largest angle ($90^\circ$), it must be the longest side of the right-angled triangle.

Let's review the options:

• (A) Smallest: Incorrect. The hypotenuse is the longest side.
• (B) Equal: Incorrect. The hypotenuse is generally not equal to the other sides, and it's always longer than each of the other sides.
• (C) Longest: Correct. The hypotenuse is the side opposite the $90^\circ$ angle, which is the largest angle, making the hypotenuse the longest side.
• (D) None of the above: Incorrect, as (C) is correct.

The correct option is (C) Longest.

Let the triangle be $\triangle ABC$. Let $AD$ be the altitude from vertex $A$ to the opposite side $BC$. The point $D$ lies on $BC$.

Since $AD$ is an altitude, it is perpendicular to $BC$. Therefore, $\angle ADB = 90^\circ$ and $\angle ADC = 90^\circ$.

The problem states that the altitude $AD$ bisects the opposite side $BC$. This means that the point $D$ is the midpoint of $BC$. Therefore, $BD = DC$.

Now, consider the two triangles formed by the altitude, $\triangle ABD$ and $\triangle ACD$.

We have the following information:

1. $BD = DC$

2. $\angle ADB = \angle ADC = 90^\circ$

3. $AD = AD$

Using the SAS (Side-Angle-Side) congruence criterion, we can say that $\triangle ABD$ is congruent to $\triangle ACD$. The equal angle ($\angle ADB = \angle ADC$) is included between the two equal sides ($BD$ and $AD$ in $\triangle ABD$, and $DC$ and $AD$ in $\triangle ACD$).

$\triangle ABD \cong \triangle ACD \quad \text{(By SAS congruence criterion)}$

By CPCT (Corresponding Parts of Congruent Triangles), the corresponding sides of congruent triangles are equal.

The side $AB$ in $\triangle ABD$ corresponds to the side $AC$ in $\triangle ACD$.

Therefore, $AB = AC$.

A triangle with two equal sides is defined as an isosceles triangle.

In an isosceles triangle, the altitude from the vertex angle bisects the base (the opposite side) and is also the angle bisector of the vertex angle and the median to the base.

If the triangle were equilateral, all three sides would be equal, which is a special case of an isosceles triangle ($AB=AC=BC$). In an equilateral triangle, the altitude from any vertex bisects the opposite side. So, an equilateral triangle also satisfies this condition.

However, a triangle where only two sides are equal (isosceles but not equilateral) also satisfies this condition for the altitude drawn from the vertex between the two equal sides.

Therefore, the general classification of the triangle is isosceles.

Let's check the options:

• (A) Scalene: A scalene triangle has all sides of different lengths. If $AB=AC$, it cannot be scalene unless $AB=AC$ is not true.
• (B) Isosceles: This is consistent with our finding that $AB = AC$.
• (C) Equilateral: While equilateral triangles satisfy this property, the condition $AB=AC$ does not guarantee that $AB=AC=BC$. So, it is not necessarily equilateral, but it must be isosceles.
• (D) Right-angled: A right-angled triangle is defined by having a $90^\circ$ angle. This condition does not require any angle to be $90^\circ$. For example, an isosceles triangle with angles $50^\circ, 65^\circ, 65^\circ$ satisfies this altitude property from the $50^\circ$ vertex, but it is not right-angled.

The most accurate description based on the given condition is that the triangle is isosceles.

The correct option is (B) Isosceles.

We are given two triangles, $\triangle ABC$ and $\triangle DEF$.

We are provided with the following information about their corresponding parts:

1. Side $AB$ of $\triangle ABC$ is equal to side $DE$ of $\triangle DEF$.

2. Angle $\angle B$ of $\triangle ABC$ is equal to angle $\angle E$ of $\triangle DEF$.

3. Side $BC$ of $\triangle ABC$ is equal to side $EF$ of $\triangle DEF$.

We need to determine the congruence criterion based on these three pieces of information.

Let's examine the arrangement of the equal parts in each triangle:

• In $\triangle ABC$, the angle $\angle B$ is included between the sides $AB$ and $BC$.
• In $\triangle DEF$, the angle $\angle E$ is included between the sides $DE$ and $EF$.

The given conditions show that two sides and the included angle of $\triangle ABC$ are equal to the corresponding two sides and the included angle of $\triangle DEF$.

This specific arrangement corresponds to the SAS (Side-Angle-Side) congruence criterion.

The SAS congruence criterion states that if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.

Therefore, $\triangle ABC$ is congruent to $\triangle DEF$ by the SAS criterion.

$\triangle ABC \cong \triangle DEF \quad \text{(By SAS congruence criterion)}$

Let's quickly consider why the other options are not applicable:

• SSS: Requires three pairs of corresponding sides to be equal. We are given only two pairs of equal sides.
• ASA: Requires two pairs of corresponding angles and the included side to be equal. We are given one pair of equal angles and two pairs of equal sides.
• RHS: Applies specifically to right-angled triangles, and requires the hypotenuse and one side to be equal. We are not told that the triangles are right-angled, and the given sides might not be the hypotenuse.

Based on the provided information ($AB=DE$, $\angle B = \angle E$, and $BC=EF$), the triangles are congruent by the SAS criterion.

The correct option is (C) SAS.

Assertion (A): If two triangles are congruent, then their areas are equal.

Two triangles are congruent if they are exact copies of each other. This means they can be placed one upon the other so that they coincide perfectly. If two figures occupy the exact same space, their areas must be equal. Therefore, Assertion (A) is true.

Reason (R): Congruent figures have the same shape and size.

This is the fundamental definition of congruent figures. Congruence implies both identical shape and identical dimensions (size). Therefore, Reason (R) is true.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that congruent triangles have equal areas. Reason (R) states that congruent figures have the same shape and size. The fact that congruent figures have the same size is precisely why they must occupy the same amount of space, leading to equal areas. The concept of having the "same size" for a two-dimensional figure is directly related to its area.

Therefore, Reason (R) correctly explains why Assertion (A) is true.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Assertion (A): If the angles of a triangle are $60^\circ, 60^\circ, 60^\circ$, it is an equilateral triangle.

A triangle with all three angles equal is called an equiangular triangle. A property of triangles states that if all angles are equal, then the sides opposite to these angles are also equal. Since all three angles are $60^\circ$ (and thus equal), the sides opposite to them must also be equal. A triangle with all sides equal is an equilateral triangle. The sum of the angles ($60^\circ + 60^\circ + 60^\circ = 180^\circ$) is valid for a triangle. Therefore, Assertion (A) is true.

Reason (R): In an equilateral triangle, all angles are equal to $60^\circ$, and all sides are equal.

An equilateral triangle is defined as a triangle with all three sides equal. A direct consequence of all sides being equal is that the angles opposite to these sides are also equal. Since the sum of angles in any triangle is $180^\circ$, if all three angles are equal, each angle must measure $\frac{180^\circ}{3} = 60^\circ$. Therefore, the statement that in an equilateral triangle, all angles are equal to $60^\circ$ and all sides are equal is the definition and a key property of equilateral triangles. Reason (R) is true.

Now let's assess if Reason (R) correctly explains Assertion (A).

Assertion (A) states that having $60^\circ, 60^\circ, 60^\circ$ angles makes a triangle equilateral. Reason (R) describes the characteristics of an equilateral triangle, including that all its angles are $60^\circ$ and all its sides are equal. Reason (R) essentially provides the definition and properties of an equilateral triangle, which confirms the statement made in Assertion (A). The fact that an equilateral triangle has all angles equal to $60^\circ$ is the fundamental link proving why a triangle with $60^\circ$ angles must be equilateral (because being equiangular and being equilateral are equivalent properties of a triangle). Thus, Reason (R) provides the basis for the truth of Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Let's match each type of triangle from Column A with its corresponding property in Column B:

• (i) Equilateral Triangle: By definition, an equilateral triangle has all three sides equal. The corresponding property in Column B is (c) All sides are equal. So, (i) matches with (c).
• (ii) Isosceles Triangle: By definition, an isosceles triangle has at least two sides equal. The corresponding property in Column B that best fits this is (d) Two sides are equal. So, (ii) matches with (d).
• (iii) Right-angled Triangle: By definition, a right-angled triangle is a triangle in which one of the angles is a right angle ($90^\circ$). The corresponding property in Column B is (a) One angle is $90^\circ$. So, (iii) matches with (a).
• (iv) Scalene Triangle: By definition, a scalene triangle has all three sides of different lengths. Consequently, all its angles are also different. The corresponding property in Column B is (b) All sides are different. So, (iv) matches with (b).

Based on the matches, we have the following pairs:

(i) - (c)

(ii) - (d)

(iii) - (a)

(iv) - (b)

Now let's compare this with the given options:

• (A) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b) - This matches our derived pairs.
• (B) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b) - Incorrect. (ii) does not match with (a), (iii) does not match with (d).
• (C) (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b) - Incorrect. (i) does not match with (a), (ii) does not match with (c), (iii) does not match with (d).
• (D) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c) - Incorrect. (i) does not match with (b), (iv) does not match with (c).

The correct option is the one that reflects the matching pairs (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

The correct option is (A) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

In this case study, we are given two triangles, $\triangle ABC$ and $\triangle XYZ$, representing bridge supports.

The civil engineer measures the sides and finds that the corresponding sides are equal:

$AB = XY$

$BC = YZ$

$CA = ZX$

We need to determine the congruence criterion that confirms the two triangles are congruent based on these equal side lengths.

The congruence criterion that uses the equality of three corresponding sides of two triangles is the SSS (Side-Side-Side) criterion.

The SSS Congruence Criterion states that if the three sides of one triangle are equal to the three corresponding sides of another triangle, then the two triangles are congruent.

Since all three pairs of corresponding sides ($AB$ and $XY$, $BC$ and $YZ$, $CA$ and $ZX$) are equal, the triangles $\triangle ABC$ and $\triangle XYZ$ satisfy the SSS congruence criterion.

Therefore, $\triangle ABC \cong \triangle XYZ$ by the SSS criterion.

Let's briefly consider the other options:

• SAS (Side-Angle-Side): Requires two sides and the included angle. This criterion cannot be applied here as we are not given any information about the angles.
• ASA (Angle-Side-Angle): Requires two angles and the included side. This criterion also cannot be applied as we are not given any information about the angles.
• RHS (Right-angle-Hypotenuse-Side): Applies specifically to right-angled triangles and requires the hypotenuse and one other side to be equal. We are not given that the triangles are right-angled, and the criterion specifically involves a right angle and the hypotenuse, which is not the information provided.

Thus, based on the given information that all three corresponding sides are equal, the congruence criterion used is SSS.

The correct option is (C) SSS.

We are asked to identify the congruence criterion based on the condition that two angles and the included side of one triangle are equal to two angles and the included side of another triangle.

Let's review the standard triangle congruence criteria:

• SSS (Side-Side-Side): All three corresponding sides are equal.
• SAS (Side-Angle-Side): Two corresponding sides and the included angle between them are equal.
• ASA (Angle-Side-Angle): Two corresponding angles and the included side between them are equal.
• AAS (Angle-Angle-Side): Two corresponding angles and a non-included side are equal.
• RHS (Right-angle-Hypotenuse-Side): In right-angled triangles, the hypotenuse and one side are equal.

The condition described in the question specifically mentions "two angles and the included side". This perfectly matches the description of the ASA congruence criterion.

The ASA criterion states that if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent.

Let's consider why the other options are not correct based on the given information:

• (B) AAS: This criterion involves two angles and a non-included side, which is different from the given condition (included side).
• (C) SAS: This criterion involves two sides and the included angle, which is also different from the given condition (two angles and the included side).
• (D) SSS: This criterion involves three sides, which is not the information provided.

Therefore, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, the triangles are congruent by the ASA criterion.

The correct option is (A) ASA.

In any triangle, there is a direct relationship between the lengths of the sides and the measures of the angles opposite to them.

The property states that the side opposite the largest angle is the longest side, and conversely, the angle opposite the longest side is the largest angle.

Similarly, the side opposite the smallest angle is the smallest side, and conversely, the angle opposite the smallest side is the smallest angle.

The question asks about the angle opposite to the smallest side. Based on this property, the angle opposite the smallest side is the smallest angle in the triangle.

Let's consider the options:

• (A) Largest: Incorrect. The largest angle is opposite the largest side.
• (B) Smallest: Correct. This aligns with the property.
• (C) Right: Incorrect. A right angle is $90^\circ$. The smallest angle in a triangle can be acute (less than $90^\circ$). Only in a right-angled triangle, the other two angles are acute, and the smallest angle among the three would be opposite the smallest leg, but it is not necessarily a right angle.
• (D) Obtuse: Incorrect. An obtuse angle is greater than $90^\circ$. In any triangle, at least two angles must be acute, and the smallest angle is always acute (less than $90^\circ$) unless it's an equilateral triangle where all angles are $60^\circ$.

Therefore, the angle opposite to the smallest side in a triangle is the smallest angle.

The correct option is (B) Smallest.

Let the triangle be $\triangle ABC$, with side lengths $a, b, c$ opposite to vertices $A, B, C$ respectively.

Let the altitudes from vertices $A, B, C$ to the opposite sides $BC, AC, AB$ be $h_a, h_b, h_c$ respectively.

The area of the triangle ($\text{Area}$) can be calculated using any side and its corresponding altitude. The formula for the area is:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude}$

Using the different sides as the base, we have:

$\text{Area} = \frac{1}{2} a h_a$

$\text{Area} = \frac{1}{2} b h_b$

$\text{Area} = \frac{1}{2} c h_c$

We are given that the altitudes of the triangle are equal. Let this common length be $h$. So, $h_a = h_b = h_c = h$.

Substituting this into the area formulas, we get:

$\text{Area} = \frac{1}{2} a h$

$\text{Area} = \frac{1}{2} b h$

$\text{Area} = \frac{1}{2} c h$

Since the area of the triangle is the same regardless of which base and altitude are used, we can equate these expressions:

$\frac{1}{2} a h = \frac{1}{2} b h = \frac{1}{2} c h$

In a non-degenerate triangle, the altitude $h$ must be a positive value ($h > 0$). Also, the factor $\frac{1}{2}$ is a non-zero constant. We can divide the entire equation by $\frac{1}{2} h$:

$\frac{\cancel{\frac{1}{2}} a \cancel{h}}{\cancel{\frac{1}{2}} \cancel{h}} = \frac{\cancel{\frac{1}{2}} b \cancel{h}}{\cancel{\frac{1}{2}} \cancel{h}} = \frac{\cancel{\frac{1}{2}} c \cancel{h}}{\cancel{\frac{1}{2}} \cancel{h}}$

This simplifies to:

$a = b = c$

This means that all three sides of the triangle are equal in length.

By definition, a triangle with all three sides equal is an equilateral triangle.

Therefore, if the altitudes of a triangle are equal, the triangle must be equilateral.

Let's evaluate the given options:

• (A) Right-angled: A right-angled triangle has one angle of $90^\circ$. The altitudes are the two legs and a segment from the right angle to the hypotenuse. These altitudes are generally not equal unless the triangle is degenerate or not right-angled.
• (B) Isosceles: An isosceles triangle has two equal sides. The altitudes to these two equal sides are equal. However, the altitude to the third side (the base) is generally different. For all three altitudes to be equal, all three sides must be equal.
• (C) Equilateral: An equilateral triangle has all sides equal. In an equilateral triangle, all altitudes are also equal. This matches our conclusion.
• (D) Scalene: A scalene triangle has all sides of different lengths. Consequently, its altitudes must also have different lengths.

The condition that the altitudes are equal is a characteristic property of equilateral triangles.

The correct option is (C) Equilateral.

In $\triangle PQR$, we are given the measures of two angles:

$\angle P = 40^\circ$

$\angle Q = 70^\circ$

We need to find the measure of the third angle, $\angle R$. The sum of the angles in any triangle is $180^\circ$.

So, $\angle P + \angle Q + \angle R = 180^\circ$

Substituting the given values, we have:

$40^\circ + 70^\circ + \angle R = 180^\circ$

$110^\circ + \angle R = 180^\circ$

$\angle R = 180^\circ - 110^\circ$

$\angle R = 70^\circ$

Now we have the measures of all three angles in $\triangle PQR$:

$\angle P = 40^\circ$

$\angle Q = 70^\circ$

$\angle R = 70^\circ$

In any triangle, the side opposite to the largest angle is the longest side, and the side opposite to the smallest angle is the shortest side.

Comparing the angles, we see that $70^\circ > 40^\circ$. The largest angles are $\angle Q$ and $\angle R$, both measuring $70^\circ$. The smallest angle is $\angle P$, measuring $40^\circ$.

The side opposite $\angle P$ is $QR$.

The side opposite $\angle Q$ is $PR$.

The side opposite $\angle R$ is $PQ$.

Since $\angle Q$ and $\angle R$ are the largest angles (both $70^\circ$), the sides opposite to them, $PR$ and $PQ$, are the longest sides. Since $\angle Q = \angle R$, the sides opposite them are equal in length, i.e., $PR = PQ$.

The side opposite the smallest angle $\angle P$ ($40^\circ$) is $QR$, which is the shortest side.

Thus, the longest sides are $PR$ and $PQ$, and they are equal in length. Among the given options, both PR and PQ are listed.

The options are (A) PR, (B) QR, (C) PQ, (D) Cannot be determined.

QR is the shortest side, so (B) is incorrect.

Both (A) PR and (C) PQ are names of the longest sides. Assuming the question intends to ask for one of the longest sides, either (A) or (C) would be correct. Since (A) is listed first and is a correct identification of a longest side, we select (A).

The correct option is (A) PR.

Given: In $\triangle ABC$, let $AD$ be the perpendicular from vertex $A$ to the opposite side $BC$. This means $AD \perp BC$, so $\angle ADB = \angle ADC = 90^\circ$. The perpendicular $AD$ also bisects the vertex angle $\angle BAC$, which means $\angle BAD = \angle CAD$.

To Determine: The type of $\triangle ABC$.

Consider the two triangles formed by the perpendicular, $\triangle ABD$ and $\triangle ACD$.

We have the following information about these two triangles:

1. $\angle BAD = \angle CAD$

2. $AD = AD$

3. $\angle ADB = \angle ADC = 90^\circ$

Based on the ASA (Angle-Side-Angle) congruence criterion, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle, then the triangles are congruent.

In $\triangle ABD$ and $\triangle ACD$, we have Angle ($\angle BAD$) - Side ($AD$) - Angle ($\angle ADB$) in $\triangle ABD$ corresponding to Angle ($\angle CAD$) - Side ($AD$) - Angle ($\angle ADC$) in $\triangle ACD$. Note that the side $AD$ is included between the equal angles $\angle BAD$ and $\angle ADB$ in $\triangle ABD$, and between the equal angles $\angle CAD$ and $\angle ADC$ in $\triangle ACD$.

Therefore, $\triangle ABD \cong \triangle ACD$ by the ASA congruence criterion.

Since the two triangles are congruent, their corresponding parts are equal by CPCT (Corresponding Parts of Congruent Triangles).

The side $AB$ in $\triangle ABD$ corresponds to the side $AC$ in $\triangle ACD$ (as they are opposite to the equal angles $\angle ADB$ and $\angle ADC$).

A triangle with two equal sides is defined as an isosceles triangle.

Therefore, if the perpendicular from a vertex to the opposite side of a triangle bisects the vertex angle, the triangle is an isosceles triangle.

Let's examine the given options:

• (A) Scalene: A scalene triangle has all sides of different lengths. Since we found $AB = AC$, the triangle cannot be scalene unless $A, B, C$ are collinear, which is not the case for a triangle.
• (B) Isosceles: An isosceles triangle has at least two equal sides. Our conclusion $AB = AC$ means the triangle is isosceles.
• (C) Equilateral: An equilateral triangle is a special case of an isosceles triangle where all three sides are equal ($AB=AC=BC$). While an equilateral triangle satisfies the given condition (the altitude from any vertex bisects the vertex angle), the condition only guarantees that $AB=AC$, not necessarily that $AB=AC=BC$. Thus, the triangle is not necessarily equilateral, but it must be isosceles.
• (D) Right-angled: A right-angled triangle has one angle equal to $90^\circ$. The given condition does not imply that any of the angles $\angle A, \angle B,$ or $\angle C$ must be $90^\circ$. For example, an isosceles triangle with angles $70^\circ, 55^\circ, 55^\circ$ satisfies the condition for the altitude from the vertex angle ($70^\circ$), but it is not right-angled.

The most accurate classification based on the given property is isosceles.

The correct option is (B) Isosceles.

We are asked to identify the condition under which two right-angled triangles are congruent by the RHS criterion.

The abbreviation RHS stands for Right-angle, Hypotenuse, Side.

This is a specific congruence criterion applicable only to right-angled triangles.

The RHS Congruence Criterion states that if in two right-angled triangles, the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle, then the two triangles are congruent.

Let's analyze the given options in the context of the RHS criterion:

• (A) Hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle. This statement exactly matches the definition of the RHS congruence criterion. Since the triangles are already stated to be right-angled, the 'R' (Right-angle) part is met. The condition then requires equality of the Hypotenuse ('H') and one other Side ('S'). This option correctly describes the RHS criterion.
• (B) Hypotenuse and one acute angle of one triangle are equal to the hypotenuse and one acute angle of the other triangle. This condition involves a hypotenuse and an angle. In a right-angled triangle, knowing the hypotenuse and one acute angle determines the other acute angle (since they sum to $90^\circ$) and the length of the sides (using trigonometry). While this condition does lead to congruence, it is typically covered by the AAS or ASA criterion when applied to right triangles (knowing the hypotenuse and one acute angle means you know two angles and a non-included side, or potentially the included side if the angle is adjacent to a leg). This is not the RHS criterion.
• (C) All sides are equal. If all sides of two triangles are equal, they are congruent by the SSS criterion. While this applies to right-angled triangles as well (e.g., two congruent $3-4-5$ triangles), it is the SSS criterion, not specifically the RHS criterion, although SSS implies RHS in right triangles with corresponding sides. The question asks for the condition for congruence *by RHS criterion*.
• (D) All angles are equal. If all angles of two triangles are equal, they are similar (equiangular triangles), but not necessarily congruent. For example, a right triangle with angles $90^\circ, 30^\circ, 60^\circ$ is similar to any other right triangle with the same angles, but they are only congruent if a corresponding side is also equal (which then leads to ASA or AAS). This is not a congruence criterion on its own.

The condition that specifically defines congruence by the RHS criterion is the equality of the hypotenuse and one side in two right-angled triangles.

The correct option is (A) Hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle.

The statement "In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side" is a fundamental geometric principle.

Let the side lengths of a triangle be $a$, $b$, and $c$. The inequality states:

$a + b > c$

$a + c > b$

$b + c > a$

This principle is known as the Triangle Inequality Theorem or simply the Triangle Inequality.

Let's examine the given options:

• (A) Pythagorean inequality: This relates to the Pythagorean theorem ($a^2 + b^2 = c^2$) for right-angled triangles. The Pythagorean inequality states that in any triangle, the square of one side is less than, equal to, or greater than the sum of the squares of the other two sides, depending on whether the angle opposite the first side is acute, right, or obtuse. This is different from the sum of the lengths of the sides.
• (B) Triangle inequality: This is the correct name for the property described.
• (C) Cauchy-Schwarz inequality: This is an inequality used in various fields of mathematics, particularly linear algebra and analysis, relating sums of products of numbers or integrals of products of functions. It is not specifically the inequality concerning the side lengths of a triangle.
• (D) Angle sum property: This property states that the sum of the interior angles of a triangle is $180^\circ$. It relates angles, not the lengths of the sides as described in the question.

Therefore, the statement describes the Triangle inequality.

The correct option is (B) Triangle.

We are given an isosceles triangle $\triangle ABC$ where side $AB$ is equal to side $AC$. This means that $A$ is the vertex angle, and $BC$ is the base. Let's analyze each given statement:

(A) $\angle B = \angle C$: This is a fundamental property of an isosceles triangle. The angles opposite the equal sides are equal. Since $AB = AC$, the angle opposite $AB$ ($\angle C$) is equal to the angle opposite $AC$ ($\angle B$). This statement is True.

(B) The median from A to BC is perpendicular to BC: In an isosceles triangle, the median drawn from the vertex angle (the angle included between the two equal sides) to the base is also the altitude to the base and the angle bisector of the vertex angle. If the median from $A$ to $BC$ is the altitude, it is perpendicular to $BC$. This statement is True.

(C) The altitude from A to BC bisects $\angle A$: As mentioned in the explanation for (B), in an isosceles triangle with $AB=AC$, the altitude from $A$ to $BC$ is also the angle bisector of $\angle A$. This statement is True.

(D) The angle bisector of $\angle B$ is equal to the angle bisector of $\angle C$: This statement refers to the lengths of the angle bisectors. In a triangle, equal angles have equal angle bisectors. Since $\angle B = \angle C$ in this isosceles triangle, their angle bisectors must be equal in length. This statement is True.

(E) All angles are equal to $60^\circ$: This is the defining property of an equilateral triangle. While an equilateral triangle is a special type of isosceles triangle (where all three sides are equal), the condition $AB=AC$ alone does not imply that $BC$ is also equal to $AB$ or $AC$. For example, an isosceles triangle can have angles $50^\circ, 65^\circ, 65^\circ$. This statement is False for a general isosceles triangle.

Based on the analysis, the properties that hold true for an isosceles triangle with $AB=AC$ are (A), (B), (C), and (D).

The correct options are (A), (B), (C), (D).

In any triangle, there is a direct relationship between the lengths of the sides and the measures of the angles opposite to them.

The property states: If two sides of a triangle are unequal, then the angle opposite the longer side is greater than the angle opposite the shorter side.

Conversely, if two angles of a triangle are unequal, then the side opposite the larger angle is longer than the side opposite the smaller angle.

Let $\triangle ABC$ be a triangle with sides $a, b, c$ opposite to vertices $A, B, C$ respectively.

If, for example, $a > b$, then the angle opposite side $a$ ($\angle A$) is greater than the angle opposite side $b$ ($\angle B$). That is, $\angle A > \angle B$.

The question asks about the angle opposite the longer side when two sides are unequal. According to the property, the angle opposite the longer side is greater than the angle opposite the shorter side.

Let's examine the options:

• (A) Smaller: Incorrect. The angle opposite the longer side is larger.
• (B) Equal: This would happen if the sides were equal (isosceles triangle), not unequal.
• (C) Greater: Correct. This matches the property.
• (D) Double: This is not a guaranteed relationship between the angles.

Therefore, if two sides of a triangle are unequal, the angle opposite to the longer side is greater than the angle opposite to the shorter side.

The correct option is (C) Greater.

The question asks for the meaning of the ASA congruence criterion for triangles.

The abbreviation ASA stands for Angle-Side-Angle.

The ASA Congruence Criterion states that if two angles and the included side (the side between the two angles) of one triangle are equal to the corresponding two angles and the included side of another triangle, then the two triangles are congruent.

Let's look at the given options:

• (A) Side, Side, Angle: This corresponds to SSA, which is NOT a general congruence criterion (it leads to the ambiguous case).
• (B) Angle, Side, Angle: This matches the expansion of ASA, where the side is included between the two angles.
• (C) Angle, Angle, Side: This corresponds to AAS, which is another congruence criterion (where the side is not included).
• (D) Side, Angle, Side: This corresponds to SAS, where the angle is included between the two sides.

Therefore, congruent by ASA criterion means Angle, Side, Angle.

The correct option is (B) Angle, Side, Angle.

Let the triangle be $\triangle ABC$. Consider the side $BC$. Let $D$ be the midpoint of $BC$. The perpendicular bisector of side $BC$ is the line that passes through $D$ and is perpendicular to $BC$.

The problem states that this perpendicular bisector passes through the opposite vertex, which is vertex $A$.

This means that vertex $A$ lies on the perpendicular bisector of the segment $BC$.

A fundamental property of the perpendicular bisector is that any point on the perpendicular bisector of a line segment is equidistant from the endpoints of the segment.

Since vertex $A$ lies on the perpendicular bisector of $BC$, the distance from $A$ to $B$ must be equal to the distance from $A$ to $C$.

A triangle in which two sides are equal in length is defined as an isosceles triangle.

Therefore, if the perpendicular bisector of a side of a triangle passes through the opposite vertex, the triangle must have the two sides adjacent to that vertex equal, making it an isosceles triangle.

Conversely, in an isosceles triangle $\triangle ABC$ with $AB=AC$, the median from $A$ to the midpoint $D$ of $BC$ is also the altitude from $A$ to $BC$. The line $AD$ is thus perpendicular to $BC$ and passes through the midpoint $D$. This line $AD$ is precisely the perpendicular bisector of $BC$, and it passes through vertex $A$. This shows the "if and only if" part of the statement.

Note that an equilateral triangle is a special case of an isosceles triangle where all three sides are equal. In an equilateral triangle, the perpendicular bisector of any side passes through the opposite vertex.

Let's check the options:

• (A) Scalene: A scalene triangle has all sides unequal. This property ($AB=AC$) does not hold.
• (B) Isosceles or Equilateral: As shown, the property holds for isosceles triangles, and equilateral triangles are a subset of isosceles triangles. This option correctly captures the condition.
• (C) Right-angled: In a right-angled triangle, the circumcenter (intersection of perpendicular bisectors) is the midpoint of the hypotenuse. Only if the triangle is also isosceles (a right isosceles triangle) would the perpendicular bisector of one leg pass through the opposite vertex (the right angle). But this is a specific case of isosceles, not all right-angled triangles.
• (D) Obtuse-angled: Similar to right-angled triangles, this property does not hold for all obtuse-angled triangles.

The correct classification is isosceles (which includes equilateral). The option "Isosceles or Equilateral" is the most accurate choice among the given options.

The correct option is (B) Isosceles or Equilateral.

We are given two triangles, $\triangle PQR$ and $\triangle STU$, with the following equal corresponding parts:

1. $\angle P = \angle S$ (Angle)

2. $\angle Q = \angle T$ (Angle)

3. $QR = TU$ (Side)

We need to determine the congruence criterion that applies here.

Let's examine the positions of the given angles and side:

• We have two pairs of equal angles ($\angle P = \angle S$ and $\angle Q = \angle T$).
• The equal side is $QR$ in $\triangle PQR$ and $TU$ in $\triangle STU$.

In $\triangle PQR$, the side $QR$ is opposite to angle $\angle P$. It is not included between angles $\angle P$ and $\angle Q$.

In $\triangle STU$, the side $TU$ is opposite to angle $\angle S$. It is not included between angles $\angle S$ and $\angle T$.

The given conditions involve two angles and a non-included side.

This configuration matches the AAS (Angle-Angle-Side) congruence criterion.

The AAS congruence criterion states that if two angles and a non-included side of one triangle are equal to two angles and the corresponding non-included side of another triangle, then the triangles are congruent.

In this case, we have $\angle P = \angle S$, $\angle Q = \angle T$, and the side $QR$ (opposite $\angle P$) is equal to the side $TU$ (opposite $\angle S$). This fits the AAS criterion.

Alternatively, since we know two angles, we can find the third angle using the angle sum property: $\angle R = 180^\circ - (\angle P + \angle Q)$ and $\angle U = 180^\circ - (\angle S + \angle T)$. Since $\angle P = \angle S$ and $\angle Q = \angle T$, it follows that $\angle R = \angle U$. Now we have $\angle Q = \angle T$, $\angle R = \angle U$, and the included side between them is $QR$ in $\triangle PQR$ and $TU$ in $\triangle STU$. Wait, the given side $QR$ is NOT included between $\angle Q$ and $\angle R$. It is included between $\angle Q$ and $\angle R$ only if the vertices are listed in the order $Q, R$. The side $QR$ is opposite to angle $P$. The side included between $\angle Q$ and $\angle R$ is $QR$. Let's re-read the problem statement and the given conditions carefully.

Given: $\angle P = \angle S$, $\angle Q = \angle T$, and $QR = TU$.

In $\triangle PQR$, the side $QR$ is opposite $\angle P$ and adjacent to $\angle Q$ and $\angle R$.

In $\triangle STU$, the side $TU$ is opposite $\angle S$ and adjacent to $\angle T$ and $\angle U$.

So, we have Angle ($\angle Q$) - Angle ($\angle P$) - Side ($QR$) in $\triangle PQR$ corresponding to Angle ($\angle T$) - Angle ($\angle S$) - Side ($TU$) in $\triangle STU$. The side is opposite to $\angle P$ and $\angle S$.

This is indeed the AAS criterion (Angle-Angle-Side, where the side is opposite one of the angles).

Let's check the options again:

• (A) ASA: Requires Angle-Included Side-Angle. We have two angles, but the side is not included between them.
• (B) AAS: Requires Angle-Angle-Side (non-included). We have $\angle Q$, $\angle P$, and the side $QR$ is opposite $\angle P$. And $\angle T$, $\angle S$, and the side $TU$ is opposite $\angle S$. This fits.
• (C) SAS: Requires Side-Included Angle-Side. We only have one side given.
• (D) RHS: Requires Right-angle-Hypotenuse-Side. We are not given right triangles.

Thus, the triangles are congruent by the AAS criterion.

The correct option is (B) AAS.

Two figures are said to be congruent if they have exactly the same shape and the same size. If two figures are congruent, one can be placed directly on top of the other and they will coincide perfectly.

Let's analyze each statement:

(A) Two circles are congruent if they have the same radius.

The shape of all circles is the same. The size of a circle is determined by its radius. If two circles have the same radius, they have the same size, and thus they are exact copies of each other. They can be made to coincide. This statement is True.

(B) Two squares are congruent if they have the same area.

The shape of all squares is the same. The size of a square is determined by its side length, say $s$. The area of a square is given by the formula $A = s^2$. If two squares have the same area, $A_1 = A_2$, then $s_1^2 = s_2^2$. Since side lengths must be positive, taking the square root gives $s_1 = s_2$. Thus, if two squares have the same area, their side lengths are equal. If their side lengths are equal, they are exact copies and are congruent. This statement is True.

(C) Two rectangles are congruent if they have the same perimeter.

Rectangles have the same general shape (four right angles), but their proportions can vary. The perimeter of a rectangle with length $l$ and width $w$ is $P = 2(l+w)$. Two different rectangles can have the same perimeter but different dimensions. For example, a rectangle with length 5 and width 1 has perimeter $2(5+1) = 12$. A rectangle with length 4 and width 2 has perimeter $2(4+2) = 12$. These two rectangles have the same perimeter but different lengths and widths, so they are not congruent. This statement is False.

(D) Two triangles are congruent if their corresponding angles are equal.

If the corresponding angles of two triangles are equal, the triangles are said to be similar. Similar triangles have the same shape but can have different sizes. For example, an equilateral triangle with side length 1 has angles $60^\circ, 60^\circ, 60^\circ$. An equilateral triangle with side length 10 also has angles $60^\circ, 60^\circ, 60^\circ$. Their corresponding angles are equal, but they are clearly not congruent as they have different side lengths. This statement is False; corresponding angles being equal is a condition for similarity, not congruence.

Based on the analysis, both statement (A) and statement (B) are correct statements regarding the congruence of the respective figures. However, in a multiple-choice question, there is usually only one intended correct answer. Option (A) provides the congruence condition based on the most fundamental size parameter of a circle (its radius). Option (B) provides the condition based on a derived property (area), although for squares, area uniquely determines the side length. Given typical mathematical convention in such questions, the statement using the most direct size defining parameter is often the preferred answer when multiple options appear correct based on derived properties. Therefore, (A) is the most likely intended correct answer.

The correct option is (A) Two circles are congruent if they have the same radius.

Given: In $\triangle DEF$, $\angle D = 30^\circ$ and $\angle E = 60^\circ$.

To Find: The shortest side of $\triangle DEF$.

In any triangle, the sum of the interior angles is $180^\circ$.

In $\triangle DEF$, we have:

$\angle D + \angle E + \angle F = 180^\circ$

Substitute the given angle measures:

$30^\circ + 60^\circ + \angle F = 180^\circ$

$90^\circ + \angle F = 180^\circ$

$\angle F = 180^\circ - 90^\circ$

$\angle F = 90^\circ$

Now we have the measures of all three angles of $\triangle DEF$:

$\angle D = 30^\circ$

$\angle E = 60^\circ$

$\angle F = 90^\circ$

The angles are $30^\circ$, $60^\circ$, and $90^\circ$. The smallest angle is $\angle D$, which measures $30^\circ$.

In a triangle, the side opposite to the smallest angle is the shortest side.

The side opposite to vertex $D$ (and thus opposite to angle $\angle D$) is the side $EF$.

Therefore, the side $EF$ is the shortest side of $\triangle DEF$.

The side opposite $\angle E$ ($60^\circ$) is $DF$.

The side opposite $\angle F$ ($90^\circ$) is $DE$ (which is the hypotenuse, the longest side in a right-angled triangle). This triangle is a right-angled triangle since $\angle F = 90^\circ$.

Comparing the sides: $EF < DF < DE$.

The shortest side is $EF$.

The correct option is (B) EF.

Given:

In $\triangle LMN$, the lengths of the sides are:

$LM = 5$ cm

$MN = 6$ cm

$NL = 7$ cm

To Find:

The largest angle in $\triangle LMN$.

Solution:

In any triangle, there is a relationship between the lengths of the sides and the measures of the angles opposite to them.

The property states that the angle opposite to the longest side is the largest angle in the triangle.

Let's compare the given side lengths:

$LM = 5$ cm

$MN = 6$ cm

$NL = 7$ cm

Arranging the side lengths in ascending order:

$5 \text{ cm} < 6 \text{ cm} < 7 \text{ cm}$

$LM < MN < NL$

The longest side is $NL$ (or $LN$), which has a length of 7 cm.

Now, we need to identify the angle opposite to the side $NL$. In $\triangle LMN$, the side $NL$ is opposite to the vertex $M$, and hence opposite to the angle $\angle M$.

According to the property, the angle opposite the longest side is the largest angle.

Since $NL$ is the longest side, the angle opposite to $NL$, which is $\angle M$, must be the largest angle in $\triangle LMN$.

The shortest side is $LM$, opposite to $\angle N$, so $\angle N$ is the smallest angle.

The side $MN$ is opposite to $\angle L$, so $\angle L$ is the middle-sized angle.

Thus, the largest angle is $\angle M$.

Let's review the options:

• (A) $\angle L$: Opposite to side $MN$ (6 cm).
• (B) $\angle M$: Opposite to side $NL$ (7 cm).
• (C) $\angle N$: Opposite to side $LM$ (5 cm).
• (D) Cannot be determined without angle measures: Incorrect, as the relationship between side lengths and opposite angles allows us to determine the relative sizes of the angles.

Comparing the side lengths, $NL$ is the longest side. The angle opposite to $NL$ is $\angle M$. Therefore, $\angle M$ is the largest angle.

The correct option is (B) $\angle M$.

We are asked to identify the valid congruence criteria for triangles from the given options.

Let's examine each option:

(A) AAA (Angle-Angle-Angle): This criterion states that if the three angles of one triangle are equal to the three corresponding angles of another triangle, then the triangles are similar, but not necessarily congruent. Similar triangles have the same shape but can have different sizes. For example, an equilateral triangle with side length 1 and an equilateral triangle with side length 5 both have angles $60^\circ, 60^\circ, 60^\circ$, but they are not congruent. Thus, AAA is not a criterion for congruence.

(B) SAS (Side-Angle-Side): This criterion states that if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent. This is a valid congruence criterion.

(C) SSA (Side-Side-Angle): This criterion states that if two sides and a non-included angle of one triangle are equal to the corresponding two sides and a non-included angle of another triangle, the triangles are not necessarily congruent. This situation is known as the ambiguous case, where two different triangles can sometimes be formed with the given information. Thus, SSA is not a general criterion for congruence.

(D) RHS (Right angle-Hypotenuse-Side): This criterion states that if in two right-angled triangles, the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle, then the triangles are congruent. This is a valid congruence criterion, specific to right-angled triangles.

Based on the analysis, the criteria from the list that are valid for proving triangle congruence are SAS and RHS.

The correct options are (B) SAS and (D) RHS.

Let the triangle be $\triangle ABC$. Let the lengths of the sides opposite to vertices $A, B, C$ be $a, b, c$ respectively. Let the lengths of the altitudes from vertices $A, B, C$ to the opposite sides be $h_a, h_b, h_c$ respectively.

The area of the triangle ($\text{Area}$) can be expressed using any side as the base and its corresponding altitude:

$\text{Area} = \frac{1}{2} \times \text{side} \times \text{altitude to that side}$

So, we have:

$\text{Area} = \frac{1}{2} a h_a$

$\text{Area} = \frac{1}{2} b h_b$

$\text{Area} = \frac{1}{2} c h_c$}

We are given that the three altitudes are equal, i.e., $h_a = h_b = h_c$. Let's denote this common length by $h$.

So, $h_a = h$, $h_b = h$, $h_c = h$.

Equating the expressions for the area:

$\frac{1}{2} a h = \frac{1}{2} b h = \frac{1}{2} c h$

Since $h$ must be greater than $0$ for a non-degenerate triangle, and $\frac{1}{2}$ is a non-zero constant, we can divide the entire equality by $\frac{1}{2} h$:

$\frac{\cancel{\frac{1}{2}} a \cancel{h}}{\cancel{\frac{1}{2}} \cancel{h}} = \frac{\cancel{\frac{1}{2}} b \cancel{h}}{\cancel{\frac{1}{2}} \cancel{h}} = \frac{\cancel{\frac{1}{2}} c \cancel{h}}{\cancel{\frac{1}{2}} \cancel{h}}$

This simplifies to:

$a = b = c$

This result shows that if the three altitudes of a triangle are equal, then the three sides of the triangle must also be equal.

A triangle with all three sides of equal length is defined as an equilateral triangle.

Let's consider the given options:

• (A) Right-angled: A right-angled triangle has one $90^\circ$ angle. The altitudes to the legs are the legs themselves. The altitude to the hypotenuse is usually different from the legs unless it's a degenerate case. Altitudes are generally not equal.
• (B) Isosceles: An isosceles triangle has two equal sides. The altitudes to these two equal sides are equal. The altitude to the third side is generally different, unless the third side is also equal, making it equilateral.
• (C) Equilateral: An equilateral triangle has all sides equal. It is a property of equilateral triangles that all three altitudes are equal in length.
• (D) Scalene: A scalene triangle has all sides of different lengths. Consequently, its altitudes must also be of different lengths.

The condition that the three altitudes of a triangle are equal uniquely identifies the triangle as equilateral.

The correct option is (C) Equilateral.

We are asked to explain what the AAS congruence criterion means.

The abbreviation AAS stands for Angle-Angle-Side.

The AAS Congruence Criterion states that if two angles and a non-included side (a side that is not between the two angles) of one triangle are equal to the corresponding two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.

Let's examine the given options:

• (A) Angle, Angle, Side (non-included): This description exactly matches the AAS criterion.
• (B) Angle, Side, Angle (included side): This describes the ASA congruence criterion, where the side is included between the two angles.
• (C) Side, Side, Angle (non-included): This describes the SSA condition, which is generally NOT a valid congruence criterion (the ambiguous case).
• (D) Side, Angle, Side (included angle): This describes the SAS congruence criterion, where the angle is included between the two sides.

Therefore, congruent by the AAS criterion means that two angles and a non-included side are equal in both triangles.

The correct option is (A) Angle, Angle, Side (non-included).

Let $\triangle ABC$ be an isosceles triangle with two equal sides $AB$ and $AC$. The side $BC$ is the base.

Let $AD$ be the median from the vertex $A$ to the base $BC$. A median connects a vertex to the midpoint of the opposite side. Therefore, $D$ is the midpoint of $BC$, which means $BD = DC$.

We want to determine the relationship between the median $AD$ and the base $BC$.

Given:

$\triangle ABC$ is isosceles with $AB = AC$.

$AD$ is the median to $BC$, so $D$ is the midpoint of $BC$, and $BD = DC$.

To Determine:

The relationship between $AD$ and $BC$.

Proof:

Consider $\triangle ABD$ and $\triangle ACD$.

We have:

$AB = AC$

$BD = CD$

$AD = AD$

By the SSS (Side-Side-Side) congruence criterion, since all three pairs of corresponding sides are equal, $\triangle ABD$ is congruent to $\triangle ACD$.

$\triangle ABD \cong \triangle ACD \quad \text{(By SSS congruence criterion)}$

By CPCT (Corresponding Parts of Congruent Triangles), the corresponding angles of congruent triangles are equal.

The angles $\angle ADB$ and $\angle ADC$ are corresponding angles in $\triangle ABD$ and $\triangle ACD$ respectively (they are opposite the equal sides $AB$ and $AC$).

$\angle$ADB = $\angle$ADC

The angles $\angle ADB$ and $\angle ADC$ form a linear pair on the straight line $BC$. Therefore, their sum is $180^\circ$.

$\angle$ADB + $\angle$ADC = $180^\circ$

Since $\angle ADB = \angle ADC$, we can substitute $\angle ADC$ with $\angle ADB$ in the equation:

$\angle$ADB + $\angle$ADB = $180^\circ$

$2 \angle$ADB = $180^\circ$

$\angle$ADB = $\frac{180^\circ}{2}$

$\angle$ADB = $90^\circ$

Since $\angle ADB = 90^\circ$, the line segment $AD$ is perpendicular to the line segment $BC$.

$AD \perp BC$

Thus, the median from the vertex angle to the base of an isosceles triangle is always perpendicular to the base.

Let's examine the given options:

• (A) Parallel: Incorrect. The median intersects the base, so it cannot be parallel to it (unless it's a degenerate triangle).
• (B) Perpendicular: Correct. As shown by the proof, the median to the base forms a $90^\circ$ angle with the base.
• (C) Equal in length: The median's length is generally not equal to the base length.
• (D) Double: The median's length is not double the base length.

The median of an isosceles triangle to the base is always perpendicular to the base.

The correct option is (B) Perpendicular.

Given:

In $\triangle XYZ$:

To Find:

The equal sides of $\triangle XYZ$.

Solution:

In any triangle, there is a relationship between the measures of the angles and the lengths of the sides opposite to them.

The property states that the sides opposite to equal angles are equal in length.

Comparing the given angle measures, we observe that $\angle Y = \angle Z = 40^\circ$.

Since $\angle Y = \angle Z$, the sides opposite to these angles must be equal.

The side opposite to $\angle Y$ is the side connecting vertices $X$ and $Z$, which is $XZ$.

The side opposite to $\angle Z$ is the side connecting vertices $X$ and $Y$, which is $XY$.

Therefore, because $\angle Y = \angle Z$, we can conclude that the sides opposite to them are equal:

The triangle is an isosceles triangle with $XY = XZ$. Note that $\angle X = 100^\circ$ is the largest angle, so the side opposite it ($YZ$) is the longest side.

Let's examine the options:

• (A) XY = YZ: Incorrect, as $YZ$ is opposite the largest angle ($100^\circ$), while $XY$ is opposite an angle of $40^\circ$.
• (B) XZ = YZ: Incorrect, similar to (A).
• (C) XY = XZ: Correct, as these sides are opposite the equal angles $\angle Z$ and $\angle Y$.
• (D) All sides are equal: Incorrect, as the angles are not all equal ($100^\circ, 40^\circ, 40^\circ$). An equilateral triangle has all angles equal to $60^\circ$.

Thus, the sides $XY$ and $XZ$ are equal.

The correct option is (C) XY = XZ.

This question is related to the Triangle Inequality Theorem.

The Triangle Inequality Theorem has two parts regarding the lengths of the sides of a triangle, say $a, b, c$:

1. The sum of the lengths of any two sides must be greater than the length of the third side.

$a + b > c$

$a + c > b$

$b + c > a$

2. The absolute difference between the lengths of any two sides must be less than the length of the third side.

$|a - b| < c$

$|a - c| < b$

$|b - c| < a$

We can derive the second part from the first. For example, starting with $a + c > b$, subtract $c$ from both sides: $a > b - c$. Starting with $b + c > a$, subtract $c$ from both sides: $b > a - c$, or $a - c < b$. Combining these two ($a > b-c$ and $a-c < b$) along with considering the case where sides might be swapped ($b > a-c$ and $b-c < a$), leads to the absolute difference $|a - b| < c$.

The statement in the question specifically asks about the difference between any two sides.

According to the second part of the Triangle Inequality Theorem, the difference between the lengths of any two sides of a triangle is always less than the length of the third side.

Let's examine the given options:

• (A) Greater: This describes the sum of any two sides, not the difference.
• (B) Equal: If the difference of two sides were equal to the third side, the points would be collinear, not forming a triangle.
• (C) Less: This is the correct relationship for the difference between any two sides compared to the third side.
• (D) Zero: The difference between two sides is zero only if they are equal. This does not provide a relationship with the third side.

Therefore, the difference between any two sides of a triangle is always less than the third side.

The correct option is (C) Less.

In this case study, the sculptor is comparing two triangles to ensure they are identical, which means they are congruent.

The sculptor measures "two sides and the angle *between* those sides" of the first triangle and finds that these match the "corresponding two sides and included angle" of the second triangle.

Let the two triangles be $\triangle ABC$ and $\triangle PQR$. The sculptor measures, for example, sides $AB$ and $BC$ and the angle $\angle B$ (which is the angle between $AB$ and $BC$). They find that $AB = PQ$, $BC = QR$, and $\angle B = \angle Q$.

The condition involves two sides and the angle that is located between these two sides. This angle is called the included angle.

The congruence criterion that uses two sides and the included angle is the SAS (Side-Angle-Side) criterion.

The SAS Congruence Criterion states that if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent.

Let's check the given options:

• (A) SSS: Requires three sides to be equal. This is not the information given.
• (B) ASA: Requires two angles and the included side to be equal. This is not the information given.
• (C) SAS: Requires two sides and the included angle to be equal. This perfectly matches the information given in the problem description.
• (D) AAS: Requires two angles and a non-included side to be equal. This is not the information given.

Therefore, based on the sculptor's measurements (two sides and the included angle), the triangles are congruent by the SAS criterion.

The correct option is (C) SAS.

Congruence of Figures:

Two geometrical figures are said to be congruent if they have exactly the same shape and the same size. In other words, two figures are congruent if they can be superimposed exactly on each other.

Congruent Line Segments:

Two line segments are congruent if and only if they have the same length.

Congruent Triangles:

Two triangles are said to be congruent if they have the same shape and the same size. This means that all corresponding sides and all corresponding angles of the two triangles are equal. If two triangles are congruent, one can be perfectly superimposed on the other by a sequence of rigid transformations (translation, rotation, reflection).

CPCTC:

CPCTC is an acronym that stands for Corresponding Parts of Congruent Triangles are Congruent. It is a fundamental principle used in geometry proofs. Once it has been established that two triangles are congruent using one of the congruence criteria (like SSS, SAS, ASA, AAS, or RHS), CPCTC can be used to prove that any pair of corresponding sides or corresponding angles are equal.

SSS Congruence Criterion:

The SSS (Side-Side-Side) congruence criterion states that:

If three sides of one triangle are respectively equal to the three corresponding sides of another triangle, then the two triangles are congruent.

SAS Congruence Criterion:

The SAS (Side-Angle-Side) congruence criterion states that:

If two sides and the included angle of one triangle are respectively equal to the two corresponding sides and the included angle of another triangle, then the two triangles are congruent.

The 'included angle' is the angle formed by the two sides being considered.

ASA Congruence Criterion:

The ASA (Angle-Side-Angle) congruence criterion states that:

If two angles and the included side of one triangle are respectively equal to the two corresponding angles and the included side of another triangle, then the two triangles are congruent.

The 'included side' is the side that lies between the two angles being considered.

RHS Congruence Criterion:

The RHS (Right angle-Hypotenuse-Side) congruence criterion states that:

If in two right-angled triangles, the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle, then the two triangles are congruent.

This criterion is specifically applicable to right-angled triangles.

Given the equalities of sides between $\triangle ABC$ and $\triangle PQR$:

• $AB = PQ$
• $BC = QR$
• $AC = PR$

According to the SSS (Side-Side-Side) congruence criterion, since all three sides of $\triangle ABC$ are equal to the corresponding three sides of $\triangle PQR$, the two triangles are congruent.

To write the congruence statement correctly, we need to match the corresponding vertices based on the equal sides:

• Since $AB = PQ$, vertex A corresponds to vertex P, and vertex B corresponds to vertex Q.
• Since $BC = QR$, vertex B corresponds to vertex Q, and vertex C corresponds to vertex R.
• Since $AC = PR$, vertex A corresponds to vertex P, and vertex C corresponds to vertex R.

All correspondences are consistent: A $\leftrightarrow$ P, B $\leftrightarrow$ Q, C $\leftrightarrow$ R.

Therefore, the congruence statement using the correct order of vertices is:

$\triangle ABC \cong \triangle PQR$

Given the equalities between $\triangle LMN$ and $\triangle XYZ$:

• $LM = XY$
• $\angle M = \angle Y$
• $MN = YZ$

The given information involves two sides and the angle included between them ($LM$ and $MN$ with included angle $\angle M$ in $\triangle LMN$, and $XY$ and $YZ$ with included angle $\angle Y$ in $\triangle XYZ$).

According to the SAS (Side-Angle-Side) congruence criterion, the two triangles are congruent.

To write the congruence statement correctly, we need to match the corresponding vertices:

• Since $LM = XY$, vertex L corresponds to vertex X, and vertex M corresponds to vertex Y. ($L \leftrightarrow X$, $M \leftrightarrow Y$)
• Since $\angle M = \angle Y$, vertex M corresponds to vertex Y. ($M \leftrightarrow Y$)
• Since $MN = YZ$, vertex M corresponds to vertex Y, and vertex N corresponds to vertex Z. ($M \leftrightarrow Y$, $N \leftrightarrow Z$)

Combining these, the vertex correspondences are $L \leftrightarrow X$, $M \leftrightarrow Y$, and $N \leftrightarrow Z$.

Therefore, the congruence statement using the correct order of vertices is:

$\triangle LMN \cong \triangle XYZ$

Given the equalities between $\triangle DEF$ and $\triangle STU$:

• $\angle E = \angle T$
• $EF = TU$
• $\angle F = \angle U$

The given information involves two angles and the side included between them ($\angle E$ and $\angle F$ with included side $EF$ in $\triangle DEF$, and $\angle T$ and $\angle U$ with included side $TU$ in $\triangle STU$).

According to the ASA (Angle-Side-Angle) congruence criterion, the two triangles are congruent.

To write the congruence statement correctly, we need to match the corresponding vertices:

• Since $\angle E = \angle T$, vertex E corresponds to vertex T. ($E \leftrightarrow T$)
• Since $\angle F = \angle U$, vertex F corresponds to vertex U. ($F \leftrightarrow U$)
• Since $EF = TU$, vertex E corresponds to vertex T, and vertex F corresponds to vertex U. ($E \leftrightarrow T$, $F \leftrightarrow U$)

The remaining vertex in $\triangle DEF$ is D, and the remaining vertex in $\triangle STU$ is S. Given the established correspondences $E \leftrightarrow T$ and $F \leftrightarrow U$, it follows that vertex D must correspond to vertex S ($D \leftrightarrow S$).

So, the vertex correspondences are $D \leftrightarrow S$, $E \leftrightarrow T$, and $F \leftrightarrow U$.

Therefore, the congruence statement using the correct order of vertices is:

$\triangle DEF \cong \triangle STU$

Given the information about right-angled triangles $\triangle ABC$ and $\triangle PQR$:

• $\triangle ABC$ is right-angled at B, so $\angle B = 90^\circ$.
• $\triangle PQR$ is right-angled at Q, so $\angle Q = 90^\circ$.
• $AC = PR$ (Hypotenuses are equal).
• $BC = QR$ (One pair of corresponding sides is equal).

In right-angled triangles, if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle, then the triangles are congruent. This is the RHS (Right angle - Hypotenuse - Side) congruence criterion.

Since $\angle B = \angle Q = 90^\circ$, $AC = PR$ (hypotenuses), and $BC = QR$ (sides), we can conclude that $\triangle ABC \cong \triangle PQR$ by the RHS criterion.

To write the congruence statement using the correct order of vertices, we match corresponding parts:

• The right angles are at vertices B and Q, so $B \leftrightarrow Q$.
• The hypotenuses are AC and PR. Since B corresponds to Q, the vertices opposite the right angles correspond. Vertex A is opposite $\angle B$, and vertex P is opposite $\angle Q$. So, $A \leftrightarrow P$. The remaining vertex C must correspond to the remaining vertex R. ($C \leftrightarrow R$).
• The equal sides are BC and QR. Since B corresponds to Q, this equality confirms that C corresponds to R.

The vertex correspondences are $A \leftrightarrow P$, $B \leftrightarrow Q$, and $C \leftrightarrow R$.

Therefore, the congruence statement using the correct order of vertices is:

$\triangle ABC \cong \triangle PQR$

Property to Prove: Angles opposite to equal sides of an isosceles triangle are equal.

Given:

Let $\triangle ABC$ be an isosceles triangle with two equal sides.

Let $AB = AC$.

To Prove:

The angles opposite the equal sides are equal.

We need to prove that $\angle ABC = \angle ACB$.

Construction:

Draw the angle bisector of $\angle BAC$. Let this bisector intersect side BC at point D.

Proof:

Consider the two triangles formed by the construction: $\triangle ABD$ and $\triangle ACD$.

In $\triangle ABD$ and $\triangle ACD$:

• $AB = AC$

  AB = AC

  (Given)

• $\angle BAD = \angle CAD$

  $\angle BAD = \angle CAD$

  (By construction, AD bisects $\angle BAC$)

• $AD = AD$

  AD = AD

  (Common side)

Therefore, by the SAS (Side-Angle-Side) congruence criterion,

$\triangle ABD \cong \triangle ACD$.

Since the triangles are congruent, their corresponding parts are congruent.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), the angle opposite side AD in $\triangle ABD$ ($\angle ABD$) must be equal to the angle opposite side AD in $\triangle ACD$ ($\angle ACD$).

Thus, $\angle ABD = \angle ACD$.

$\angle ABD$ is the same as $\angle ABC$, and $\angle ACD$ is the same as $\angle ACB$.

Hence, $\angle ABC = \angle ACB$.

This proves that the angles opposite to the equal sides of an isosceles triangle are equal.

Property to Prove: Sides opposite to equal angles of a triangle are equal.

Given:

Let $\triangle ABC$ be a triangle.

Let $\angle ABC = \angle ACB$.

To Prove:

The sides opposite the equal angles are equal.

We need to prove that $AB = AC$.

Construction:

Draw the angle bisector of $\angle BAC$. Let this bisector intersect side BC at point D.

Proof:

Consider the two triangles formed by the construction: $\triangle ABD$ and $\triangle ACD$.

In $\triangle ABD$ and $\triangle ACD$:

• $\angle ABD = \angle ACD$

  $\angle ABD = \angle ACD$

  (Given, $\angle ABC = \angle ACB$)

• $\angle BAD = \angle CAD$

  $\angle BAD = \angle CAD$

  (By construction, AD bisects $\angle BAC$)

• $AD = AD$

  AD = AD

  (Common side)

Therefore, by the ASA (Angle-Side-Angle) congruence criterion,

$\triangle ABD \cong \triangle ACD$.

Since the triangles are congruent, their corresponding parts are congruent.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), the side opposite $\angle ADB$ in $\triangle ABD$ ($AB$) must be equal to the side opposite $\angle ADC$ in $\triangle ACD$ ($AC$). Wait, this is incorrect application of CPCTC based on the chosen angles. Let's correct the CPCTC application.

By CPCTC, the side opposite $\angle ADB$ in $\triangle ABD$ ($AB$) corresponds to the side opposite $\angle ADC$ in $\triangle ACD$ ($AC$). Also, the side opposite $\angle ABD$ ($AD$) corresponds to the side opposite $\angle ACD$ ($AD$), which we already used. The side opposite $\angle BAD$ ($BD$) corresponds to the side opposite $\angle CAD$ ($CD$). The angles opposite side AD ($\angle ABD$ and $\angle ACD$) correspond and are equal. The angles opposite side BD and CD ($\angle BAD$ and $\angle CAD$) correspond and are equal. The angles opposite side AB and AC ($\angle ADB$ and $\angle ADC$) correspond. The sides opposite $\angle ABD$ ($AD$) and $\angle ACD$ ($AD$) correspond. The sides opposite $\angle BAD$ ($BD$) and $\angle CAD$ ($CD$) correspond. The sides opposite $\angle ADB$ ($AB$) and $\angle ADC$ ($AC$) correspond.

Since $\triangle ABD \cong \triangle ACD$, the corresponding sides are equal.

The side opposite to $\angle ABD$ (which is $\angle ABC$) in $\triangle ABD$ is $AD$. The side opposite to $\angle ACD$ (which is $\angle ACB$) in $\triangle ACD$ is $AD$. This doesn't help prove $AB = AC$.

Let's look at the correspondence from the congruence $\triangle ABD \cong \triangle ACD$. Since the angles $\angle BAD$ and $\angle CAD$ are equal, the sides opposite to them must correspond. Side opposite $\angle BAD$ is $BD$. Side opposite $\angle CAD$ is $CD$. So $BD = CD$. Since the angles $\angle ABD$ and $\angle ACD$ are equal, the sides opposite to them must correspond. Side opposite $\angle ABD$ is $AD$. Side opposite $\angle ACD$ is $AD$. This is the common side. The remaining angle in $\triangle ABD$ is $\angle ADB$. The remaining angle in $\triangle ACD$ is $\angle ADC$. Since the triangles are congruent, $\angle ADB = \angle ADC$. The sides opposite these equal angles must correspond. Side opposite $\angle ADB$ is $AB$. Side opposite $\angle ADC$ is $AC$.

Thus, by CPCTC, the sides opposite the corresponding angles $\angle ADB$ and $\angle ADC$ are equal.

Therefore, $AB = AC$.

This proves that the sides opposite to the equal angles of a triangle are equal.

Triangle Inequality Theorem:

The Triangle Inequality Theorem states that:

In any triangle, the sum of the lengths of any two sides of the triangle is greater than the length of the third side.

Let the sides of a triangle be $a$, $b$, and $c$. The theorem can be expressed by the following three inequalities:

• $a + b > c$
• $a + c > b$
• $b + c > a$

To determine if a triangle can be formed with given side lengths, we use the Triangle Inequality Theorem.

The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Given side lengths are 4 cm, 5 cm, and 10 cm. Let's check if the Triangle Inequality Theorem holds for all pairs of sides:

• Check if the sum of the first two sides is greater than the third side:

  $4 + 5 > 10$

  $9 > 10$

  This inequality is False.

• Check if the sum of the first and third sides is greater than the second side:

  $4 + 10 > 5$

  $14 > 5$

  This inequality is True.

• Check if the sum of the second and third sides is greater than the first side:

  $5 + 10 > 4$

  $15 > 4$

  This inequality is True.

For a triangle to be formed, all three inequalities must be true. In this case, the first inequality ($4 + 5 > 10$) is false because $9$ is not greater than $10$.

Conclusion:

No, a triangle cannot be formed with side lengths 4 cm, 5 cm, and 10 cm because the sum of two sides ($4 + 5 = 9$) is not greater than the third side (10).

Given Angles:

The angles of the triangle are $50^\circ$, $60^\circ$, and $70^\circ$.

Largest Angle:

Comparing the three angles, the largest angle is $70^\circ$.

Theorem:

The theorem states that in any triangle, the side opposite to the larger angle is longer. Conversely, the angle opposite to the longer side is larger.

Identifying the Longest Side:

According to the theorem, the side opposite the largest angle ($70^\circ$) will be the longest side in the triangle.

Therefore, the side opposite the angle measuring $70^\circ$ is the longest side.

Conclusion:

The longest side in the triangle with angles $50^\circ, 60^\circ,$ and $70^\circ$ is the side opposite the $70^\circ$ angle.

Given Side Lengths:

The side lengths of the triangle are 7 cm, 9 cm, and 12 cm.

Longest Side:

Comparing the three side lengths, the longest side is 12 cm.

Theorem:

The theorem states that in a triangle, the angle opposite to the longer side is larger.

Identifying the Largest Angle:

According to the theorem, the angle opposite the longest side (12 cm) will be the largest angle in the triangle.

Conclusion:

The largest angle in the triangle with side lengths 7 cm, 9 cm, and 12 cm is the angle opposite the side measuring 12 cm.

Given:

• In $\triangle ABC$, $\angle A = 40^\circ$
• $\angle B = 80^\circ$

To Find:

• Measure of $\angle C$.
• Sides in ascending order of length.

Solution:

We know that the sum of angles in a triangle is $180^\circ$.

In $\triangle ABC$, we have:

$\angle A + \angle B + \angle C = 180^\circ$

Substitute the given values:

$40^\circ + 80^\circ + \angle C = 180^\circ$

$120^\circ + \angle C = 180^\circ$

$\angle C = 180^\circ - 120^\circ$

$\angle C = 60^\circ$

Now we have the measures of all three angles:

• $\angle A = 40^\circ$
• $\angle B = 80^\circ$
• $\angle C = 60^\circ$

To arrange the sides in ascending order of length, we use the property that the side opposite the smaller angle is shorter, and the side opposite the larger angle is longer.

Let's arrange the angles in ascending order:

$40^\circ < 60^\circ < 80^\circ$

So, $\angle A < \angle C < \angle B$.

The sides opposite these angles are:

• Side opposite $\angle A$ is BC.
• Side opposite $\angle C$ is AB.
• Side opposite $\angle B$ is AC.

According to the property, the order of the sides in ascending length will be the same as the order of the angles opposite to them.

Since $\angle A < \angle C < \angle B$, the sides opposite to them will be in the same order of length:

Side opposite $\angle A$ < Side opposite $\angle C$ < Side opposite $\angle B$

$BC < AB < AC$.

Conclusion:

The measure of $\angle C$ is $60^\circ$.

The sides of the triangle in ascending order of length are BC, AB, and AC.

$BC < AB < AC$.

Given:

• In $\triangle PQR$, $PQ = 6$ cm
• $QR = 8$ cm
• $RP = 7$ cm

To Find:

Angles in descending order of measure.

Solution:

We are given the lengths of the sides of $\triangle PQR$.

The side lengths are: 6 cm, 8 cm, and 7 cm.

Let's arrange the side lengths in descending order:

$8 \text{ cm} > 7 \text{ cm} > 6 \text{ cm}$

So, the order of sides from longest to shortest is: $QR > RP > PQ$.

We use the property that in a triangle, the angle opposite to the longer side is larger.

The angles opposite to these sides are:

• Angle opposite side QR (8 cm) is $\angle P$.
• Angle opposite side RP (7 cm) is $\angle Q$.
• Angle opposite side PQ (6 cm) is $\angle R$.

According to the property, the order of the angles from largest to smallest will be the same as the order of the sides opposite to them from longest to shortest.

Since $QR > RP > PQ$, the angles opposite to them will be in the same order of measure:

Angle opposite QR > Angle opposite RP > Angle opposite PQ

$\angle P > \angle Q > \angle R$.

Conclusion:

The angles of $\triangle PQR$ in descending order of measure are $\angle P$, $\angle Q$, and $\angle R$.

$\angle P > \angle Q > \angle R$.

Minimum Number of Equal Parts:

To prove two triangles congruent, the minimum number of corresponding parts (sides or angles) required to be equal is three.

Sufficiency of Any Three Equal Parts:

No, any three equal parts are not sufficient to prove two triangles congruent. The specific combination and relative position of the three equal parts are crucial.

Explanation with Example (SSA/ASS Case):

Consider the case where two sides and a non-included angle (often referred to as SSA or ASS) of one triangle are equal to the corresponding two sides and non-included angle of another triangle.

Let's say in $\triangle ABC$ and $\triangle DEF$, we have:

• $AB = DE$ (Side)
• $BC = EF$ (Side)
• $\angle A = \angle D$ (Non-included Angle, opposite to side BC and EF respectively)

Even with these three corresponding parts equal, the triangles may not be congruent. This is known as the ambiguous case of congruence.

Example:

Imagine you are given a side length, an angle, and another side length that is NOT between the angle and the first side. For instance, side $AB = 6$ cm, angle $\angle A = 30^\circ$, and side $BC = 4$ cm.

If you fix side AB and angle A, you can draw a ray from A at $30^\circ$. Now, from point B, you need to draw an arc with radius 4 cm to intersect this ray. In some cases, this arc can intersect the ray at two different points (say C$_1$ and C$_2$), creating two different triangles ($\triangle ABC_1$ and $\triangle ABC_2$).

Both $\triangle ABC_1$ and $\triangle ABC_2$ would satisfy the conditions:

• $AB$ is common to both.
• $\angle A$ is common to both.
• $BC_1 = BC_2 = 4$ cm.

However, $\triangle ABC_1$ and $\triangle ABC_2$ are clearly not congruent (they have different angle measures at B and C, and different lengths for the third side $AC$).

This example shows that having two sides and a non-included angle equal (three parts) is not enough to guarantee congruence.

The sufficient congruence criteria (SSS, SAS, ASA, AAS, RHS) ensure that the equal parts are in a specific, rigid arrangement that fixes the shape and size of the triangle uniquely. Any other combination of three parts is not sufficient.

Given:

• In $\triangle ABC$ and $\triangle XYZ$:
• $\angle A = \angle X$
• $\angle B = \angle Y$
• $AB = XY$

Analysis:

We are given that two angles ($\angle A$ and $\angle B$) and the side included between them ($AB$) in $\triangle ABC$ are equal to the corresponding two angles ($\angle X$ and $\angle Y$) and the included side ($XY$) in $\triangle XYZ$.

In $\triangle ABC$, the side $AB$ is included between the angles $\angle A$ and $\angle B$.

In $\triangle XYZ$, the side $XY$ is included between the angles $\angle X$ and $\angle Y$.

Congruence Criterion:

The congruence criterion that uses two angles and the included side is the ASA (Angle-Side-Angle) congruence criterion.

Conclusion:

Since $\angle A = \angle X$, $AB = XY$, and $\angle B = \angle Y$, the triangles $\triangle ABC$ and $\triangle XYZ$ are congruent by the ASA congruence criterion.

The congruence statement is $\triangle ABC \cong \triangle XYZ$.

No, two triangles with equal corresponding angles are not necessarily congruent.

Explanation:

Having all corresponding angles equal means that the two triangles have the same shape, but not necessarily the same size. Triangles with the same shape are called similar triangles. Congruent triangles have both the same shape and the same size.

The Angle-Angle-Angle (AAA) criterion proves similarity, not congruence. For congruence, at least one corresponding side must also be equal (as seen in ASA or AAS criteria, which can be derived from ASA using the angle sum property).

Example:

Consider two equilateral triangles, $\triangle ABC$ and $\triangle PQR$.

In $\triangle ABC$, let each side measure 2 cm. Since it's equilateral, all angles are $60^\circ$. So, $\angle A = \angle B = \angle C = 60^\circ$.

In $\triangle PQR$, let each side measure 4 cm. Since it's equilateral, all angles are also $60^\circ$. So, $\angle P = \angle Q = \angle R = 60^\circ$.

Comparing the corresponding angles:

• $\angle A = \angle P = 60^\circ$
• $\angle B = \angle Q = 60^\circ$
• $\angle C = \angle R = 60^\circ$

All corresponding angles are equal.

Comparing the corresponding sides:

• $AB = 2$ cm, $PQ = 4$ cm ($AB \neq PQ$)
• $BC = 2$ cm, $QR = 4$ cm ($BC \neq QR$)
• $AC = 2$ cm, $PR = 4$ cm ($AC \neq PR$)

The corresponding sides are not equal.

Although $\triangle ABC$ and $\triangle PQR$ have equal corresponding angles, they are not congruent because they have different side lengths (and thus different sizes). $\triangle PQR$ is simply a larger version of $\triangle ABC$. They are similar, but not congruent. A diagram would clearly show two equilateral triangles of different sizes, highlighting their identical shape but distinct size.

Given:

• In $\triangle ABC$, AD is the median to BC.
• $AB > AC$

To Compare:

$\angle ADC$ and $\angle ADB$.

Solution:

Since AD is the median to BC, point D is the midpoint of BC.

Therefore, $BD = CD$.

Consider $\triangle ABD$ and $\triangle ACD$.

We are given the side lengths:

• AB $>$ AC

  (Given)

• BD = CD

  (Definition of median)

• AD is a common side to both triangles.

We use the theorem which states that in any triangle, the angle opposite to the longer side is larger.

In $\triangle ABD$, the side opposite to $\angle ADB$ is AB.

In $\triangle ACD$, the side opposite to $\angle ADC$ is AC.

Since we are given that $AB > AC$, according to the theorem, the angle opposite side AB must be greater than the angle opposite side AC.

Therefore, the angle opposite AB in $\triangle ABD$ ($\angle ADB$) is greater than the angle opposite AC in $\triangle ACD$ ($\angle ADC$).

So, we have:

$\angle ADB > \angle ADC$

Alternate Approach (Using Exterior Angle Property - Though not strictly necessary here, good for confirmation):

Consider $\triangle ABD$. $\angle ADC$ is an exterior angle to $\triangle ABD$ if we extend BD to C.

The exterior angle $\angle ADC$ is equal to the sum of the two opposite interior angles of $\triangle ABD$.

$\angle ADC = \angle BAD + \angle ABD$

This means $\angle ADC > \angle ABD$. This approach doesn't directly compare $\angle ADC$ and $\angle ADB$.

Let's stick to the direct application of the side-angle relationship.

Conclusion:

Given that $AB > AC$ in $\triangle ABC$ where AD is the median to BC, the angle opposite the longer side AB in $\triangle ABD$ is $\angle ADB$, and the angle opposite the shorter side AC in $\triangle ACD$ is $\angle ADC$.

Therefore, $\angle ADB$ is greater than $\angle ADC$.

$\angle ADB > \angle ADC$.

Yes, in a right-angled triangle, the hypotenuse is indeed the longest side.

Justification using Inequality Properties:

Consider a right-angled triangle, say $\triangle ABC$, where $\angle B = 90^\circ$.

According to the angle sum property of a triangle:

$\angle A + \angle B + \angle C = 180^\circ$

Substituting $\angle B = 90^\circ$:

$\angle A + 90^\circ + \angle C = 180^\circ$

$\angle A + \angle C = 180^\circ - 90^\circ$

$\angle A + \angle C = 90^\circ$

Since $\angle A$ and $\angle C$ are angles in a triangle, they must be positive measures ($\angle A > 0^\circ$ and $\angle C > 0^\circ$).

From $\angle A + \angle C = 90^\circ$, it follows that:

• $\angle A < 90^\circ$
• $\angle C < 90^\circ$

Comparing the angles in $\triangle ABC$:

The angles are $\angle A$, $\angle B$ ($90^\circ$), and $\angle C$.

Since $\angle A < 90^\circ$ and $\angle C < 90^\circ$, the angle $\angle B = 90^\circ$ is the largest angle in the right-angled triangle.

Now, we use the inequality property that states: In any triangle, the side opposite to the larger angle is longer.

The side opposite the largest angle ($\angle B = 90^\circ$) is the hypotenuse (AC).

The sides opposite the other two angles ($\angle A$ and $\angle C$) are BC and AB, respectively. These are the legs of the right triangle.

Since $\angle B$ is the largest angle, the side opposite to it, AC (the hypotenuse), must be longer than the sides opposite to $\angle A$ (BC) and $\angle C$ (AB).

Thus, $AC > BC$ and $AC > AB$.

Conclusion:

The hypotenuse is the side opposite the largest angle ($90^\circ$) in a right-angled triangle. By the theorem relating angles and opposite sides, the hypotenuse is the longest side.

The four main congruence criteria for triangles are methods used to prove that two triangles are congruent, meaning they have the same size and shape.

1. SSS (Side-Side-Side) Congruence Criterion:

This criterion states that if all three sides of one triangle are equal in length to the corresponding three sides of another triangle, then the two triangles are congruent.

Conditions:

In $\triangle ABC$ and $\triangle DEF$, if:

• $AB = DE$
• $BC = EF$
• $AC = DF$

Then, $\triangle ABC \cong \triangle DEF$ by the SSS criterion.

Diagram:

Draw two triangles, ABC and DEF. Mark sides AB and DE with the same symbol (e.g., a single dash), sides BC and EF with another symbol (e.g., a double dash), and sides AC and DF with a third symbol (e.g., a triple dash) to indicate their equality.

2. SAS (Side-Angle-Side) Congruence Criterion:

This criterion states that if two sides and the angle included between them of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the two triangles are congruent.

Conditions:

In $\triangle ABC$ and $\triangle DEF$, if:

• $AB = DE$ (Side)
• $\angle B = \angle E$ (Included Angle)
• $BC = EF$ (Side)

Then, $\triangle ABC \cong \triangle DEF$ by the SAS criterion.

Diagram:

Draw two triangles, ABC and DEF. Mark sides AB and DE with the same symbol, sides BC and EF with another symbol. Mark the angles included between these sides, $\angle B$ and $\angle E$, with the same arc symbol.

3. ASA (Angle-Side-Angle) Congruence Criterion:

This criterion states that if two angles and the side included between them of one triangle are equal to the corresponding two angles and the included side of another triangle, then the two triangles are congruent.

Conditions:

In $\triangle ABC$ and $\triangle DEF$, if:

• $\angle B = \angle E$ (Angle)
• $BC = EF$ (Included Side)
• $\angle C = \angle F$ (Angle)

Then, $\triangle ABC \cong \triangle DEF$ by the ASA criterion.

Diagram:

Draw two triangles, ABC and DEF. Mark angles $\angle B$ and $\angle E$ with the same arc symbol, and angles $\angle C$ and $\angle F$ with another arc symbol. Mark the sides included between these angles, BC and EF, with the same dash symbol.

4. RHS (Right angle-Hypotenuse-Side) Congruence Criterion:

This criterion is specific to right-angled triangles. It states that if the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse and one side of another right-angled triangle, then the two triangles are congruent.

Conditions:

In right-angled $\triangle ABC$ (at $\angle B = 90^\circ$) and right-angled $\triangle DEF$ (at $\angle E = 90^\circ$), if:

• $\angle B = \angle E = 90^\circ$ (Right angle)
• $AC = DF$ (Hypotenuse)
• $BC = EF$ (One side)

Then, $\triangle ABC \cong \triangle DEF$ by the RHS criterion. Note that either pair of corresponding legs can be the equal side (e.g., $AB = DE$ instead of $BC = EF$).

Diagram:

Draw two right-angled triangles, ABC (right angle at B) and DEF (right angle at E). Mark the right angles with the standard square symbol. Mark the hypotenuses AC and DF with the same dash symbol. Mark one pair of corresponding legs, say BC and EF, with another dash symbol.

Given:

In quadrilateral ABCD, the diagonals AC and BD are drawn.

• AD = BC

• BD = AC

To Prove:

• $\triangle ABD \cong \triangle BAC$
• $\angle DAB = \angle CBA$

Proof:

Consider triangles $\triangle ABD$ and $\triangle BAC$.

In $\triangle ABD$ and $\triangle BAC$:

• AB = BA

  (Common side)

• BD = AC

  (Given)

• AD = BC

  (Given)

Since the three sides of $\triangle ABD$ are equal to the corresponding three sides of $\triangle BAC$, the two triangles are congruent by the SSS congruence criterion.

Thus, $\triangle ABD \cong \triangle BAC$.

Now that the triangles are proven congruent, their corresponding parts are equal by CPCTC.

The angle $\angle DAB$ in $\triangle ABD$ is opposite to the side BD.

The angle $\angle CBA$ in $\triangle BAC$ is opposite to the side AC.

Since $BD = AC$ (corresponding sides from the congruence), their opposite angles must also be equal.

Therefore, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent):

Hence, it is proven that $\triangle ABD \cong \triangle BAC$ and $\angle DAB = \angle CBA$.

Given:

AB is a line segment.

P and Q are points on opposite sides of AB such that:

• PA = PB
• QA = QB

To Prove:

The line PQ is the perpendicular bisector of AB.

Let M be the point where PQ intersects AB. We need to prove that:

• AM = MB (PQ bisects AB)
• $\angle PMA = 90^\circ$ (PQ is perpendicular to AB)

Proof:

Consider $\triangle PAQ$ and $\triangle PBQ$.

In $\triangle PAQ$ and $\triangle PBQ$:

• PA = PB

  (Given)

• QA = QB

  (Given)

• PQ = PQ

  (Common side)

Therefore, by the SSS congruence criterion, $\triangle PAQ \cong \triangle PBQ$.

Since the triangles are congruent, their corresponding angles are equal by CPCTC.

Let M be the point of intersection of PQ and AB. Then $\angle APM$ and $\angle BPQ$ are the same angles as $\angle APQ$ and $\angle BPQ$ respectively, considering the line segment PM as part of PQ.

So, $\angle APM = \angle BPM$.

Now consider $\triangle APM$ and $\triangle BPM$.

In $\triangle APM$ and $\triangle BPM$:

• PA = PB

  (Given)

• $\angle APM = \angle BPM$

  (Proved above)

• PM = PM

  (Common side)

Therefore, by the SAS congruence criterion, $\triangle APM \cong \triangle BPM$.

Since the triangles $\triangle APM$ and $\triangle BPM$ are congruent, their corresponding parts are equal by CPCTC.

This proves that M is the midpoint of AB, so PQ bisects AB.

Also by CPCTC:

Angles $\angle PMA$ and $\angle PMB$ form a linear pair on the line segment AB.

So, $\angle PMA + \angle PMB = 180^\circ$ (Linear Pair Axiom).

Since $\angle PMA = \angle PMB$, we can substitute $\angle PMA$ for $\angle PMB$:

$\angle PMA + \angle PMA = 180^\circ$

$2 \angle PMA = 180^\circ$

$\angle PMA = \frac{180^\circ}{2}$

$\angle PMA = 90^\circ$

This means $\angle PMB = 90^\circ$ as well.

Since $\angle PMA = 90^\circ$, the line PQ is perpendicular to AB.

Since PQ bisects AB at M (AM = BM) and is perpendicular to AB ($\angle PMA = 90^\circ$), the line PQ is the perpendicular bisector of AB.

Hence Proved.

Given:

In the given figure:

AC = AE

AB = AD

$\angle BAD = \angle EAC$

To Prove:

BC = DE

Proof:

We are given that $\angle BAD = \angle EAC$.

Consider the angle $\angle DAC$. It is common to both $\angle BAC$ and $\angle DAE$.

Add $\angle DAC$ to both sides of the given equality $\angle BAD = \angle EAC$:

$\angle BAD + \angle DAC = \angle EAC + \angle DAC$

From the figure, we observe that:

Substituting these into the equation above, we get:

Now consider $\triangle ABC$ and $\triangle ADE$.

In $\triangle ABC$ and $\triangle ADE$:

• AB = AD

  (Given)

• $\angle BAC = \angle DAE$

  (From (i))

• AC = AE

  (Given)

We have two sides and the included angle of $\triangle ABC$ equal to the corresponding two sides and the included angle of $\triangle ADE$.

Therefore, by the SAS congruence criterion:

$\triangle ABC \cong \triangle ADE$ (By SAS)

Since the triangles are congruent, their corresponding parts are equal by CPCTC.

The side BC in $\triangle ABC$ corresponds to the side DE in $\triangle ADE$.

Thus, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent):

Hence Proved.

Triangles have certain inequality properties that relate the lengths of their sides and the measures of their angles.

(a) Sum of Any Two Sides is Greater Than the Third Side (Triangle Inequality Theorem):

Statement: In any triangle, the sum of the lengths of any two sides is greater than the length of the third side.

Explanation: This property is a fundamental consequence of the fact that the shortest distance between two points is a straight line. If we consider any two vertices of a triangle, say A and B, the straight line segment AB represents the shortest path between them. Any other path from A to B that goes through the third vertex, say C (i.e., following the path AC + CB), must be longer than the direct path AB, unless A, B, and C are collinear (which is not the case for a triangle). This holds true for all three pairs of sides.

Let the side lengths of a triangle be $a, b,$ and $c$. The property can be expressed as:

• $a + b > c$
• $a + c > b$
• $b + c > a$

Numerical Example:

Consider side lengths 3 cm, 4 cm, and 5 cm.

• $3 + 4 = 7 > 5$ (True)
• $3 + 5 = 8 > 4$ (True)
• $4 + 5 = 9 > 3$ (True)

Since the sum of any two sides is greater than the third side, a triangle can be formed with these side lengths.

Now consider side lengths 3 cm, 4 cm, and 8 cm.

• $3 + 4 = 7$
• $7 > 8$ (False)

Since $3 + 4$ is not greater than 8, a triangle cannot be formed with these side lengths. The condition fails.

Diagram:

Draw a triangle ABC. Label the side opposite vertex A as 'a' (BC), the side opposite vertex B as 'b' (AC), and the side opposite vertex C as 'c' (AB). Write the inequalities next to the diagram.

(b) The Angle Opposite the Longer Side is Larger:

Statement: In any triangle, the angle opposite the longer side has a greater measure than the angle opposite the shorter side.

The converse is also true: The side opposite the larger angle is longer than the side opposite the smaller angle.

Explanation: This property establishes a direct relationship between the size of an angle in a triangle and the length of the side that it "opens up to". A larger angle will necessarily have a longer side across from it, and conversely, a longer side will be opposite a larger angle. This property helps in comparing angles based on side lengths and vice versa without knowing the exact measures.

In $\triangle ABC$, if side $AB > AC$, then the angle opposite AB ($\angle C$) is greater than the angle opposite AC ($\angle B$).

If $\angle A > \angle B$, then the side opposite $\angle A$ (BC) is longer than the side opposite $\angle B$ (AC).

Numerical Example:

Consider a triangle with side lengths 7 cm, 9 cm, and 12 cm.

The side lengths in ascending order are 7 cm, 9 cm, 12 cm.

Let these sides be p, q, r where p=7, q=9, r=12. The angles opposite to these sides are $\angle P, \angle Q, \angle R$ respectively.

Since $12 > 9 > 7$, the side of length 12 cm is the longest, the side of length 9 cm is the next longest, and the side of length 7 cm is the shortest.

According to the property, the order of the angles opposite these sides will correspond to the order of the side lengths.

• The angle opposite the 12 cm side (say $\angle R$) is the largest angle.
• The angle opposite the 9 cm side (say $\angle Q$) is the next largest angle.
• The angle opposite the 7 cm side (say $\angle P$) is the smallest angle.

So, $\angle R > \angle Q > \angle P$.

Diagram:

Draw a triangle ABC with clearly different side lengths (e.g., make AC visibly longer than BC, and BC visibly longer than AB). Label the sides with numerical values (e.g., AB=5, BC=7, AC=9). Indicate the angles opposite these sides. Show that the angle opposite the longest side (angle B) is the largest, and the angle opposite the shortest side (angle C) is the smallest.

Property to Prove: The perpendicular line segment from a point to a line (not on it) is the shortest of all line segments drawn from the point to the line.

Given:

Let L be a line and P be a point not on L.

Let PM be the perpendicular line segment from P to the line L, where M is on L.

Let PN be any other line segment from P to the line L, where N is on L and $N \neq M$.

To Prove:

$PM < PN$

Proof:

Consider the triangle $\triangle PMN$.

In $\triangle PMN$, we are given that PM is perpendicular to the line L.

Thus, $\angle PMN = 90^\circ$.

In any triangle, the sum of the angles is $180^\circ$.

In $\triangle PMN$:

$\angle MPN + \angle PMN + \angle PNM = 180^\circ$

Substitute $\angle PMN = 90^\circ$:

$\angle MPN + 90^\circ + \angle PNM = 180^\circ$

$\angle MPN + \angle PNM = 180^\circ - 90^\circ$

$\angle MPN + \angle PNM = 90^\circ$

Since the sum of the other two angles is $90^\circ$, each of these angles must be less than $90^\circ$.

So, $\angle PNM < 90^\circ$.

And $\angle MPN < 90^\circ$.

Comparing the angles in $\triangle PMN$, we have:

$\angle PMN = 90^\circ$ (The right angle)

$\angle PNM < 90^\circ$ (An acute angle)

Therefore, $\angle PMN > \angle PNM$.

Now, we use the triangle inequality property that states: The side opposite to the larger angle is longer.

The side opposite to $\angle PMN$ ($90^\circ$) is PN (the hypotenuse in the right triangle).

The side opposite to $\angle PNM$ (the acute angle) is PM (the perpendicular segment).

Since $\angle PMN > \angle PNM$, the side opposite $\angle PMN$ must be longer than the side opposite $\angle PNM$.

Thus, $PN > PM$.

Or equivalently, $PM < PN$.

This shows that the perpendicular segment PM is shorter than any other segment PN drawn from P to the line L.

Hence Proved.

Given:

In $\triangle PQR$, $PQ > PR$.

QM is the median to side PR (M is the midpoint of PR).

RN is the median to side PQ (N is the midpoint of PQ).

To Prove:

$QM > RN$

Proof:

We are given that in $\triangle PQR$, $PQ > PR$.

By the triangle inequality property, the angle opposite the longer side is larger.

Therefore, the angle opposite side PQ ($\angle PRQ$ or $\angle R$) is greater than the angle opposite side PR ($\angle PQR$ or $\angle Q$).

Let G be the centroid of $\triangle PQR$. The centroid is the point of intersection of the medians QM and RN.

The centroid G divides each median in the ratio 2:1, with the larger part towards the vertex.

So, $QG = \frac{2}{3} QM$ and $GM = \frac{1}{3} QM$.

And $RG = \frac{2}{3} RN$ and $GN = \frac{1}{3} RN$.

We want to prove $QM > RN$. This is equivalent to proving $\frac{2}{3} QM > \frac{2}{3} RN$, which is $QG > RG$.

Consider $\triangle QGR$. The sides of this triangle are QR, QG, and RG.

By the triangle inequality property, in $\triangle QGR$, the side opposite the larger angle is longer. Therefore, $QG > RG$ if and only if $\angle GRQ > \angle GQR$.

Thus, to prove $QG > RG$, we need to prove $\angle GRQ > \angle GQR$.

We know that $\angle PRQ > \angle PQR$. $\angle GRQ$ is the angle at vertex R within $\triangle QGR$, and $\angle GQR$ is the angle at vertex Q within $\triangle QGR$. These are parts of the larger angles $\angle PRQ$ and $\angle PQR$.

Let's use a proof by contradiction to show $\angle GRQ > \angle GQR$. Assume $\angle GRQ \le \angle GQR$.

Case 1: $\angle GRQ = \angle GQR$.

In $\triangle QGR$, if $\angle GRQ = \angle GQR$, then the sides opposite these angles are equal.

$QG = RG$ (Sides opposite equal angles are equal).

Since $QG = \frac{2}{3} QM$ and $RG = \frac{2}{3} RN$, $QG = RG \implies \frac{2}{3} QM = \frac{2}{3} RN \implies QM = RN$.

It is a known theorem that if the medians to two sides of a triangle are equal, then the sides to which they are drawn are equal.

So, $QM = RN \implies PR = PQ$.

But this contradicts the given condition $PQ > PR$.

Case 2: $\angle GRQ < \angle GQR$.

In $\triangle QGR$, if $\angle GRQ < \angle GQR$, then the side opposite the smaller angle is shorter than the side opposite the larger angle.

$QG < RG$ (Side opposite smaller angle is shorter).

Since $QG = \frac{2}{3} QM$ and $RG = \frac{2}{3} RN$, $QG < RG \implies \frac{2}{3} QM < \frac{2}{3} RN \implies QM < RN$.

It is a known theorem (converse of the previous) that if one median is shorter than another median, the side to which the first median is drawn is shorter than the side to which the second median is drawn.

So, $QM < RN \implies PR < PQ$.

This is consistent with the given condition $PQ > PR$. However, this assumption was based on the premise $\angle GRQ < \angle GQR$ contradicting $\angle PRQ > \angle PQR$. We need to show that $\angle GRQ < \angle GQR$ leads to $\angle PRQ < \angle PQR$, which contradicts the given.

Let's use the established fact that $\angle GRQ < \angle GQR$ implies $QM < RN$, which implies $PR < PQ$.

So, the assumption $\angle GRQ \le \angle GQR$ (which covers both $\angle GRQ = \angle GQR$ and $\angle GRQ < \angle GQR$) implies $PQ \le PR$.

However, we are given $PQ > PR$.

This is a contradiction.

Therefore, the initial assumption $\angle GRQ \le \angle GQR$ must be false.

Thus, $\angle GRQ > \angle GQR$.

Now, returning to $\triangle QGR$, since $\angle GRQ > \angle GQR$, by the angle-side inequality, the side opposite $\angle GRQ$ is greater than the side opposite $\angle GQR$.

$QG > RG$.

Substitute $QG = \frac{2}{3} QM$ and $RG = \frac{2}{3} RN$:

$\frac{2}{3} QM > \frac{2}{3} RN$

Multiplying both sides by $3/2$:

$QM > RN$

Hence Proved.

Given:

• AB is a line segment.
• P is the midpoint of AB.
• D and E are points on the same side of AB.
• $\angle BAD = \angle ABE$
• $\angle EPA = \angle DPB$

To Prove:

• $\triangle DAP \cong \triangle EBP$
• AD = BE

Proof:

We are given that $\angle EPA = \angle DPB$.

Let's add $\angle EPD$ to both sides of this equation.

$\angle EPA + \angle EPD = \angle DPB + \angle EPD$

From the figure, we can see that $\angle EPA + \angle EPD$ forms $\angle APD$.

Also, $\angle DPB + \angle EPD$ forms $\angle BPE$.

Therefore, we have:

Now consider triangles $\triangle DAP$ and $\triangle EBP$.

In $\triangle DAP$ and $\triangle EBP$:

• $\angle DAP = \angle EBP$

  (Given, since $\angle BAD = \angle ABE$ and P lies on AB)

• AP = BP

  (Given that P is the midpoint of AB)

• $\angle APD = \angle BPE$

  (From (i))

We have two angles ($\angle DAP$ and $\angle APD$ in $\triangle DAP$, and $\angle EBP$ and $\angle BPE$ in $\triangle EBP$) and the included side (AP in $\triangle DAP$, and BP in $\triangle EBP$) equal to the corresponding parts of the other triangle.

Therefore, by the ASA (Angle-Side-Angle) congruence criterion, $\triangle DAP$ is congruent to $\triangle EBP$.

$\triangle DAP \cong \triangle EBP$ (By ASA)

Since the triangles $\triangle DAP$ and $\triangle EBP$ are congruent, their corresponding parts are equal.

By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), the side AD in $\triangle DAP$ must be equal to the corresponding side BE in $\triangle EBP$.

Thus, we have:

Hence Proved.

Given:

In $\triangle ABC$:

• $\angle ACB = 90^\circ$ (right angled at C)
• M is the midpoint of the hypotenuse AB.
• C is joined to M and produced to D such that DM = CM.
• D is joined to B.

To Prove:

• (a) $\triangle AMC \cong \triangle BMD$
• (b) $\angle DBC = 90^\circ$
• (c) $\triangle DBC \cong \triangle ACB$
• (d) $CM = \frac{1}{2} AB$

Proof:

(a) Proving $\triangle AMC \cong \triangle BMD$:

Consider $\triangle AMC$ and $\triangle BMD$.

In $\triangle AMC$ and $\triangle BMD$:

• AM = BM

  (Given that M is the midpoint of AB)

• $\angle AMC = \angle BMD$

  (Vertically opposite angles)

• CM = DM

  (By construction)

By the SAS congruence criterion, $\triangle AMC \cong \triangle BMD$.

(b) Proving $\angle DBC$ is a right angle:

Since $\triangle AMC \cong \triangle BMD$ (from part a), their corresponding parts are equal by CPCTC.

• AC = BD

  (CPCTC)

• $\angle MAC = \angle MBD$

  (CPCTC)

The angles $\angle MAC$ (or $\angle CAB$) and $\angle MBD$ (or $\angle DBA$) are alternate interior angles formed by the transversal AB intersecting lines AC and BD.

Since the alternate interior angles are equal, the lines AC and BD must be parallel.

Thus, AC || BD.

Now consider BC as a transversal intersecting the parallel lines AC and BD.

The interior angles on the same side of the transversal are supplementary.

So, $\angle ACB + \angle DBC = 180^\circ$.

We are given that $\angle ACB = 90^\circ$.

Substituting this value:

$\angle DBC = 180^\circ - 90^\circ$

$\angle DBC = 90^\circ$.

Therefore, $\angle DBC$ is a right angle.

(c) Proving $\triangle DBC \cong \triangle ACB$:

Consider $\triangle DBC$ and $\triangle ACB$.

In $\triangle DBC$ and $\triangle ACB$:

• BC = CB

  (Common side)

• $\angle DBC = \angle ACB$

  ($90^\circ$ each, proved in part b and given)

• BD = AC

  (Proved in part b, from $\triangle AMC \cong \triangle BMD$ by CPCTC)

By the SAS congruence criterion, $\triangle DBC \cong \triangle ACB$.

(d) Proving CM = $\frac{1}{2}$ AB:

Since $\triangle DBC \cong \triangle ACB$ (from part c), their corresponding parts are equal by CPCTC.

The side CD in $\triangle DBC$ is opposite the right angle at B (hypotenuse of $\triangle DBC$).

The side AB in $\triangle ACB$ is opposite the right angle at C (hypotenuse of $\triangle ACB$).

Therefore, by CPCTC:

From the construction, we know that D is a point on the line CM produced such that DM = CM.

Thus, the length of the segment CD is the sum of CM and DM.

CD = CM + DM

Since DM = CM (by construction), we can write:

CD = CM + CM

CD = 2 CM

We proved that CD = AB. Substituting this into the equation above:

Dividing both sides by 2:

$CM = \frac{1}{2} AB$

Hence Proved.

Given:

In $\triangle ABC$:

• AB = AC
• BO is the bisector of $\angle ABC$.
• CO is the bisector of $\angle ACB$.

BO and CO intersect at O.

To Prove:

• OB = OC
• AO bisects $\angle BAC$ (or $\angle A$)

Proof:

Part 1: Proving OB = OC

In $\triangle ABC$, we are given that $AB = AC$.

According to the property of isosceles triangles, the angles opposite to equal sides are equal.

We are given that BO is the bisector of $\angle ABC$ and CO is the bisector of $\angle ACB$.

This means that BO divides $\angle ABC$ into two equal halves, and CO divides $\angle ACB$ into two equal halves.

$\angle OBC = \frac{1}{2}\angle ABC$

$\angle OCB = \frac{1}{2}\angle ACB$

Since $\angle ABC = \angle ACB$, multiplying both sides by $\frac{1}{2}$, we get:

Now consider $\triangle OBC$.

In $\triangle OBC$, we have shown that $\angle OBC = \angle OCB$.

According to the property that sides opposite to equal angles of a triangle are equal, the side opposite $\angle OBC$ must be equal to the side opposite $\angle OCB$.

The side opposite $\angle OBC$ is OC.

The side opposite $\angle OCB$ is OB.

Therefore, $OC = OB$.

Part 2: Proving AO bisects $\angle BAC$

Consider $\triangle ABO$ and $\triangle ACO$.

In $\triangle ABO$ and $\triangle ACO$:

• AB = AC

  (Given)

• OB = OC

  (Proved in Part 1)

• AO = AO

  (Common side)

Since the three sides of $\triangle ABO$ are equal to the corresponding three sides of $\triangle ACO$, the two triangles are congruent by the SSS congruence criterion.

Thus, $\triangle ABO \cong \triangle ACO$ (By SSS).

Since the triangles are congruent, their corresponding parts are equal by CPCTC.

The angle $\angle BAO$ in $\triangle ABO$ corresponds to the angle $\angle CAO$ in $\triangle ACO$.

This means that the line segment AO divides the angle $\angle BAC$ into two equal angles.

Therefore, AO bisects $\angle BAC$ (or $\angle A$).

Hence Proved.

We are asked to show that the quadrilateral ABED is a parallelogram based on the given information.

Given:

• In $\triangle ABC$ and $\triangle DEF$:
• $AB = DE$
• $AB \parallel DE$
• $BC = EF$
• $BC \parallel EF$
• Vertices A, B, C are joined to D, E, F respectively.

To Prove:

Quadrilateral ABED is a parallelogram.

Proof:

Consider the quadrilateral ABED. Its vertices are A, B, E, and D.

The sides of quadrilateral ABED are AB, BE, ED, and DA.

We are given the following information concerning two of these sides (which are opposite sides in ABED):

• AB = DE

  (Given)

• AB $\parallel$ DE

  (Given)

We know a property of quadrilaterals that states:

"If one pair of opposite sides of a quadrilateral is equal in length and parallel, then the quadrilateral is a parallelogram."

In quadrilateral ABED, the pair of opposite sides AB and ED (which is the same as DE) satisfies these two conditions: $AB = ED$ and $AB \parallel ED$.

Therefore, based on the property mentioned above, the quadrilateral ABED must be a parallelogram.

Hence Proved.

(Note: The information about BC=EF and BC $\parallel$ EF is relevant for proving other parts, such as quadrilateral BCFE being a parallelogram, and ultimately leading to $\triangle ABC \cong \triangle DEF$, but it is not needed to prove that ABED is a parallelogram).

Given:

In $\triangle ABC$, AD is the median to side BC.

To Prove:

$\angle BAC = 90^\circ$

Proof:

Since AD is the median to side BC, D is the midpoint of BC.

This means that BD and CD are equal in length, and each is half the length of BC.

We are given that $AD = \frac{1}{2} BC$.

Combining this with the relation $BD = CD = \frac{1}{2} BC$, we have:

Consider $\triangle ABD$.

From (i), we have $AD = BD$.

Since two sides are equal, $\triangle ABD$ is an isosceles triangle.

In an isosceles triangle, the angles opposite the equal sides are equal.

The angle opposite side AD is $\angle ABD$.

The angle opposite side BD is $\angle BAD$.

So, $\angle BAD = \angle ABD$.

Let $\angle BAD = \angle ABD = x$.

Consider $\triangle ACD$.

From (i), we have $AD = CD$.

Since two sides are equal, $\triangle ACD$ is an isosceles triangle.

The angle opposite side AD is $\angle ACD$.

The angle opposite side CD is $\angle CAD$.

So, $\angle CAD = \angle ACD$.

Let $\angle CAD = \angle ACD = y$.

Now consider the angles of the main triangle $\triangle ABC$.

$\angle BAC = \angle BAD + \angle CAD = x + y$.

$\angle ABC = \angle ABD = x$.

$\angle ACB = \angle ACD = y$.

The sum of the angles in any triangle is $180^\circ$.

In $\triangle ABC$:

$\angle BAC + \angle ABC + \angle ACB = 180^\circ$

Substitute the expressions in terms of x and y:

$(x + y) + x + y = 180^\circ$

$2x + 2y = 180^\circ$

Factor out 2:

$2(x + y) = 180^\circ$

Divide by 2:

$x + y = \frac{180^\circ}{2}$

$x + y = 90^\circ$

Since $\angle BAC = x + y$, we have:

$\angle BAC = 90^\circ$

Therefore, $\angle BAC$ is a right angle.

Hence Proved.