Triangle Properties: Angles and Sides
Triangle Properties: Angles and Sides - Angle Sum Property of a Triangle
One of the most fundamental and widely used properties of triangles in Euclidean geometry is the Angle Sum Property. This property establishes a constant relationship between the measures of the three interior angles of any triangle, regardless of its size or shape. It is a direct consequence of the Parallel Postulate or its equivalents.
Theorem Statement
The sum of the angles of a triangle is $180^\circ$.
Proof of the Angle Sum Property
To prove this theorem, we will use the properties of parallel lines and transversals, which are based on Euclid's Fifth Postulate.
Given: A triangle PQR with interior angles $\angle \text{P}$ (or $\angle \text{QPR}$), $\angle \text{Q}$ (or $\angle \text{PQR}$), and $\angle \text{R}$ (or $\angle \text{PRQ}$). Let's denote their measures as $\text{m}\angle 1, \text{m}\angle 2$, and $\text{m}\angle 3$ respectively for simplicity in the diagram.
To Prove: $\text{m}\angle 1 + \text{m}\angle 2 + \text{m}\angle 3 = 180^\circ$.
Construction: Draw a straight line XY passing through vertex P such that line XY is parallel to the side QR (XY || QR).
Proof:
XPY is a straight line.
(By construction)
Angles $\angle \text{XPQ}$, $\angle \text{QPR}$, and $\angle \text{RPY}$ are on the straight line XPY. Let $\text{m}\angle \text{XPQ} = \text{m}\angle 4$ and $\text{m}\angle \text{RPY} = \text{m}\angle 5$.
(Angles on a straight line at point P)
$\text{m}\angle \text{XPQ} + \text{m}\angle \text{QPR} + \text{m}\angle \text{RPY} = 180^\circ$
(Sum of angles on a straight line is $180^\circ$ - Linear Pair Axiom extended)
$\text{m}\angle 4 + \text{m}\angle 1 + \text{m}\angle 5 = 180^\circ$
(Substituting notation from diagram)
Line XY || Line QR and PQ is a transversal.
(By construction and Given)
$\angle \text{XPQ}$ and $\angle \text{PQR}$ are alternate interior angles.
(Definition of Alternate Interior Angles)
$\text{m}\angle \text{XPQ} = \text{m}\angle \text{PQR}$
(Alternate Interior Angles are equal when lines are parallel)
$\text{m}\angle 4 = \text{m}\angle 2$
(Substituting notation from diagram)
Line XY || Line QR and PR is a transversal.
(By construction and Given)
$\angle \text{RPY}$ and $\angle \text{PRQ}$ are alternate interior angles.
(Definition of Alternate Interior Angles)
$\text{m}\angle \text{RPY} = \text{m}\angle \text{PRQ}$
(Alternate Interior Angles are equal when lines are parallel)
$\text{m}\angle 5 = \text{m}\angle 3$
(Substituting notation from diagram)
Substitute $\text{m}\angle 4 = \text{m}\angle 2$ and $\text{m}\angle 5 = \text{m}\angle 3$ into statement $\text{m}\angle 4 + \text{m}\angle 1 + \text{m}\angle 5 = 180^\circ$.
(Substitution Principle / Common Notion 1)
$\text{m}\angle 2 + \text{m}\angle 1 + \text{m}\angle 3 = 180^\circ$
(Result of substitution)
Rearranging the terms:
$\text{m}\angle 1 + \text{m}\angle 2 + \text{m}\angle 3 = 180^\circ$
Since $\angle 1, \angle 2, \angle 3$ are the interior angles of $\triangle \text{PQR}$, the sum of the angles of the triangle is $180^\circ$.
Hence Proved.
Example
Example 1. In a $\triangle \text{ABC}$, if $\angle \text{A} = 70^\circ$ and $\angle \text{B} = 50^\circ$, find the measure of $\angle \text{C}$.
Answer:
Given: In $\triangle \text{ABC}$, $\text{m}\angle \text{A} = 70^\circ$ and $\text{m}\angle \text{B} = 50^\circ$.
To Find: The measure of $\angle \text{C}$.
Solution:
According to the Angle Sum Property of a triangle, the sum of the measures of the three interior angles of any triangle is $180^\circ$.
$\text{m}\angle \text{A} + \text{m}\angle \text{B} + \text{m}\angle \text{C} = 180^\circ$
(Angle Sum Property)
Substitute the given values of $\text{m}\angle \text{A}$ and $\text{m}\angle \text{B}$ into the equation:
$70^\circ + 50^\circ + \text{m}\angle \text{C} = 180^\circ$
Combine the known angle measures:
$120^\circ + \text{m}\angle \text{C} = 180^\circ$
To find $\text{m}\angle \text{C}$, subtract $120^\circ$ from both sides of the equation:
$\text{m}\angle \text{C} = 180^\circ - 120^\circ$
$\mathbf{m}\angle \text{C} = 60^\circ$
The measure of angle C is $60^\circ$.
Triangle Properties: Angles and Sides - Exterior Angle Property of a Triangle
In addition to the Angle Sum Property of a triangle, which deals with the relationship between the interior angles, there is another important theorem concerning the angles formed when a side of a triangle is extended. This is known as the Exterior Angle Property. It relates an angle formed outside the triangle to the interior angles that are not adjacent to it.
Theorem Statement
The Exterior Angle Property theorem states:
Theorem: If a side of a triangle is produced (extended outwards), then the exterior angle so formed is equal to the sum of the two interior opposite angles.
Proof of the Exterior Angle Property
Given: A triangle PQR. The side QR is produced to a point S, forming the exterior angle $\angle \text{PRS}$. The interior angles of $\triangle \text{PQR}$ are $\angle \text{PQR}$ ($\angle 1$), $\angle \text{QRP}$ ($\angle 2$), and $\angle \text{RPQ}$ ($\angle 3$). The exterior angle $\angle \text{PRS}$ is denoted as $\angle 4$. The two interior opposite angles to $\angle \text{PRS}$ are $\angle \text{PQR}$ ($\angle 1$) and $\angle \text{RPQ}$ ($\angle 3$).
To Prove: $\text{m}\angle 4 = \text{m}\angle 1 + \text{m}\angle 3$.
Proof:
The line segment QRS is a straight line.
(By construction, QR is produced to S)
Angles $\angle \text{QRP}$ ($\angle 2$) and $\angle \text{PRS}$ ($\angle 4$) form a linear pair on the straight line QRS at vertex R.
(Definition of Linear Pair)
$\text{m}\angle \text{QRP} + \text{m}\angle \text{PRS} = 180^\circ$
(Linear Pair Axiom)
$\text{m}\angle 2 + \text{m}\angle 4 = 180^\circ$
... (i)
In $\triangle \text{PQR}$, the sum of the interior angles is $180^\circ$.
(Angle Sum Property of a Triangle)
$\text{m}\angle \text{PQR} + \text{m}\angle \text{QRP} + \text{m}\angle \text{RPQ} = 180^\circ$
$\text{m}\angle 1 + \text{m}\angle 2 + \text{m}\angle 3 = 180^\circ$
... (ii)
From equation (i) and equation (ii), since both sums are equal to $180^\circ$, we can equate them:
$\text{m}\angle 2 + \text{m}\angle 4 = \text{m}\angle 1 + \text{m}\angle 2 + \text{m}\angle 3$
(Things equal to the same thing are equal to one another - Common Notion 1)
Subtract $\text{m}\angle 2$ from both sides of the equation:
$\mathbf{\text{m}\angle 4 = \text{m}\angle 1 + \text{m}\angle 3}$
(If equals are subtracted from equals, the remainders are equal - Common Notion 3)
Thus, the measure of the exterior angle $\angle \text{PRS}$ is equal to the sum of the measures of the two interior opposite angles $\angle \text{PQR}$ and $\angle \text{RPQ}$.
Hence Proved.
Important Consequence:
From the Exterior Angle Property ($\angle 4 = \angle 1 + \angle 3$), since $\angle 1$ and $\angle 3$ are angle measures (which are positive values for a non-degenerate triangle), it follows that the exterior angle of a triangle is always strictly greater than the measure of either of its interior opposite angles individually.
$\text{m}\angle 4 > \text{m}\angle 1$
$\text{m}\angle 4 > \text{m}\angle 3$
This consequence is sometimes useful in comparing angle measures within a triangle.
Example 1. In the figure below, find the value of $x$.
Answer:
Given: A triangle with two interior opposite angles measuring $50^\circ$ and $x^\circ$. The exterior angle corresponding to these interior opposite angles measures $120^\circ$.
To Find: The value of $x$.
Solution:
According to the Exterior Angle Property of a triangle, the measure of an exterior angle is equal to the sum of the measures of its two interior opposite angles.
Exterior Angle = Sum of Interior Opposite Angles
Substituting the given values into the property:
$120^\circ = 50^\circ + x^\circ$
To find the value of $x$, subtract $50^\circ$ from both sides of the equation:
$x^\circ = 120^\circ - 50^\circ$
$x^\circ = 70^\circ$
Therefore, the value of $x$ is $70$. The measure of the angle is $70^\circ$.
Triangle Properties: Angles and Sides - Properties related to Special Triangles (Isosceles Triangle Theorem)
We have classified triangles based on their side lengths and angle measures. Among these, the isosceles triangle, with its property of having at least two equal sides, possesses specific angle properties that are described by the Isosceles Triangle Theorem and its converse. These theorems establish a direct link between the equality of sides and the equality of angles within a triangle.
Isosceles Triangle Theorem (Base Angles Theorem)
This theorem is fundamental to understanding the properties of isosceles triangles.
Theorem Statement: Angles opposite to equal sides of an isosceles triangle are equal.
In an isosceles triangle, the two equal sides are called the legs, and the third side is called the base. The angles opposite the legs (which are the angles at the base) are called the base angles, and the angle opposite the base (the angle between the two legs) is called the vertex angle.
The theorem states that the base angles of an isosceles triangle are equal.
Proof of the Isosceles Triangle Theorem
We can prove this theorem using the concept of triangle congruence.
Given: An isosceles triangle ABC in which side AB is equal to side AC ($\text{AB} = \text{AC}$). $\angle \text{B}$ is opposite to side AC, and $\angle \text{C}$ is opposite to side AB.
To Prove: $\text{m}\angle \text{B} = \text{m}\angle \text{C}$.
Construction: Draw the angle bisector of the vertex angle $\angle \text{BAC}$ (or $\angle \text{A}$). Let this angle bisector be AD, which intersects the side BC at point D.
Proof:
Now, consider the two triangles formed by the construction, $\triangle \text{ABD}$ and $\triangle \text{ACD}$.
1. AB = AC
(Given)
2. $\angle \text{BAD} = \angle \text{CAD}$
(By Construction - AD bisects $\angle \text{BAC}$)
3. AD = AD
(Common side to both triangles)
Therefore, by the Side-Angle-Side (SAS) congruence rule:
$\triangle \text{ABD} \cong \triangle \text{ACD}$
(SAS congruence criteria)
Since the two triangles $\triangle \text{ABD}$ and $\triangle \text{ACD}$ are congruent, their corresponding parts are equal.
$\angle \text{ABD} = \angle \text{ACD}$
(Corresponding Parts of Congruent Triangles - CPCT)
This means $\text{m}\angle \text{B} = \text{m}\angle \text{C}$.
Hence, the angles opposite to the equal sides AB and AC are equal.
Hence Proved.
Furthermore, by CPCT, we also get $\text{BD} = \text{CD}$ (D is the midpoint of BC) and $\angle \text{ADB} = \angle \text{ADC}$. Since $\angle \text{ADB}$ and $\angle \text{ADC}$ form a linear pair, $\text{m}\angle \text{ADB} + \text{m}\angle \text{ADC} = 180^\circ$. As they are equal, each must be $90^\circ$. Thus, the angle bisector of the vertex angle in an isosceles triangle is also the median to the base and the altitude to the base, and is perpendicular to the base.
Converse of Isosceles Triangle Theorem
The converse of the theorem reverses the hypothesis and conclusion.
Theorem Statement: The sides opposite to equal angles of a triangle are equal.
This theorem states that if two angles in a triangle are equal, then the sides opposite those angles are also equal, meaning the triangle is isosceles (or equilateral if all three angles are equal).
Proof of the Converse of Isosceles Triangle Theorem
We can again use triangle congruence to prove the converse.
Given: A triangle ABC in which angle B is equal to angle C ($\text{m}\angle \text{B} = \text{m}\angle \text{C}$).
To Prove: AB = AC.
Construction: Draw the angle bisector AD of the vertex angle $\angle \text{BAC}$ (or $\angle \text{A}$). Let this angle bisector be AD, which intersects the side BC at point D.
Proof:
Now, consider the two triangles formed by the construction, $\triangle \text{ABD}$ and $\triangle \text{ACD}$.
1. $\angle \text{BAD} = \angle \text{CAD}$
(By Construction - AD bisects $\angle \text{BAC}$)
2. AD = AD
(Common side to both triangles)
3. $\angle \text{ABD} = \angle \text{ACD}$
(Given $\text{m}\angle \text{B} = \text{m}\angle \text{C}$, so $\text{m}\angle \text{ABD} = \text{m}\angle \text{ACD}$)
Therefore, by the Angle-Angle-Side (AAS) congruence rule:
$\triangle \text{ABD} \cong \triangle \text{ACD}$
(AAS congruence criteria)
Since the two triangles $\triangle \text{ABD}$ and $\triangle \text{ACD}$ are congruent, their corresponding parts are equal.
AB = AC
(Corresponding Parts of Congruent Triangles - CPCT)
Thus, the side opposite to angle C (AB) is equal to the side opposite to angle B (AC).
Hence Proved.
Example
Example 1. In $\triangle \text{PQR}$, $PQ = PR$ and $\angle \text{Q} = 65^\circ$. Find the measure of $\angle \text{P}$.
Answer:
Given: In $\triangle \text{PQR}$, $\text{PQ} = \text{PR}$ and $\text{m}\angle \text{Q} = 65^\circ$.
To Find: The measure of $\angle \text{P}$.
Solution:
We are given that in $\triangle \text{PQR}$, the sides PQ and PR are equal in length ($\text{PQ} = \text{PR}$). This means $\triangle \text{PQR}$ is an isosceles triangle.
According to the Isosceles Triangle Theorem, the angles opposite the equal sides are equal.
The angle opposite to side PQ is $\angle \text{R}$.
The angle opposite to side PR is $\angle \text{Q}$.
$\angle \text{R} = \angle \text{Q}$
(Angles opposite equal sides of an isosceles triangle)
Since $\text{m}\angle \text{Q} = 65^\circ$, then
$\text{m}\angle \text{R} = 65^\circ$
Now, we can use the Angle Sum Property of a triangle, which states that the sum of the interior angles of any triangle is $180^\circ$.
$\text{m}\angle \text{P} + \text{m}\angle \text{Q} + \text{m}\angle \text{R} = 180^\circ$
(Angle Sum Property)
Substitute the known values of $\text{m}\angle \text{Q}$ and $\text{m}\angle \text{R}$:
$\text{m}\angle \text{P} + 65^\circ + 65^\circ = 180^\circ$
Combine the known angle measures:
$\text{m}\angle \text{P} + 130^\circ = 180^\circ$
To find $\text{m}\angle \text{P}$, subtract $130^\circ$ from both sides:
$\text{m}\angle \text{P} = 180^\circ - 130^\circ$
$\mathbf{\text{m}\angle \text{P}} = 50^\circ$
The measure of angle P is $50^\circ$.
Triangle Properties: Angles and Sides - Side Length Properties (Sum of Lengths of Two Sides)
Beyond the angle relationships, the side lengths of a triangle also have a fundamental relationship with each other. This property, known as the Triangle Inequality Theorem, is a key principle that determines whether a set of three given lengths can actually form a triangle.
Triangle Inequality Theorem
Theorem Statement: The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
This theorem can be stated as three inequalities for a triangle with side lengths $a, b$, and $c$:
- $a + b > c$
- $a + c > b$
- $b + c > a$
All three of these conditions must be satisfied for the three lengths to form the sides of a triangle. If even one of these inequalities is false, then a triangle cannot be formed with those side lengths.
Intuitively, this makes sense. The shortest distance between two points is a straight line. The third side of the triangle represents the straight line distance between two vertices, while the sum of the other two sides represents a path from one vertex to the third via the second vertex. This indirect path must be longer than the direct path (the third side).
Proof of the Triangle Inequality Theorem
Given: A triangle ABC with sides AB, BC, and AC.
To Prove: $\text{AB} + \text{AC} > \text{BC}$. (We will prove one inequality; the other two inequalities can be proven using a similar construction and logic).
Construction: Extend the side BA to a point D such that the length of the segment AD is equal to the length of the side AC ($\text{AD} = \text{AC}$). Join point D to point C to form the line segment DC, creating a larger triangle BDC.
Proof:
Consider the triangle $\triangle \text{ADC}$ formed by the construction.
1. AD = AC
(By Construction)
Since $\triangle \text{ADC}$ has two equal sides, it is an isosceles triangle. According to the Isosceles Triangle Theorem, the angles opposite to equal sides are equal.
2. $\angle \text{ADC} = \angle \text{ACD}$
(Angles opposite equal sides in $\triangle \text{ADC}$)
Now, consider the angle $\angle \text{BCD}$. From the figure, $\angle \text{BCD}$ is the sum of $\angle \text{BCA}$ and $\angle \text{ACD}$.
3. $\angle \text{BCD} = \angle \text{BCA} + \angle \text{ACD}$
(Angle Addition Postulate)
Since $\angle \text{BCA}$ is a positive angle measure (for a non-degenerate triangle), from statement 3:
4. $\angle \text{BCD} > \angle \text{ACD}$
(Whole is greater than the part - Common Notion 5)
Substitute $\angle \text{ACD}$ with $\angle \text{ADC}$ (from statement 2) in statement 4. Note that $\angle \text{ADC}$ is the same as $\angle \text{BDC}$ as D, A, B are collinear.
5. $\angle \text{BCD} > \angle \text{BDC}$
(Substitution from 2 into 4)
Now, consider the large triangle $\triangle \text{BDC}$. We have just shown that $\angle \text{BCD}$ is greater than $\angle \text{BDC}$. A property of triangles (which will be covered in the next section on Inequalities in a Triangle) states that the side opposite the larger angle is longer than the side opposite the smaller angle.
6. The side opposite $\angle \text{BCD}$ in $\triangle \text{BDC}$ is BD.
7. The side opposite $\angle \text{BDC}$ in $\triangle \text{BDC}$ is BC.
8. BD > BC
(Side opposite larger angle is longer)
From the construction, the length of BD is the sum of the lengths of BA and AD.
9. BD = BA + AD
(Segment Addition Postulate)
We also know from construction that AD = AC.
10. Substitute BD from statement 9 into statement 8: (BA + AD) > BC.
(Substitution Principle)
11. Substitute AD = AC (from statement 1) into statement 10.
(Substitution Principle)
$\mathbf{\text{BA} + \text{AC} > \text{BC}}$
(Result of substitution)
Which is the same as $\text{AB} + \text{AC} > \text{BC}$.
Thus, the sum of the lengths of sides AB and AC is greater than the length of side BC. Similarly, by extending other sides, we can prove that $\text{AB} + \text{BC} > \text{AC}$ and $\text{BC} + \text{AC} > \text{AB}$.
Hence Proved.
Corollary: Difference of Two Sides
A direct consequence or corollary of the Triangle Inequality Theorem is the relationship between the difference of two sides and the third side.
Corollary Statement: The difference between the lengths of any two sides of a triangle is less than the length of the third side.
For a triangle with side lengths $a, b,$ and $c$, this means:
- $|a - b| < c$
- $|a - c| < b$
- $|b - c| < a$
The absolute value is used because the difference could be negative depending on which side is subtracted from which, but length is always positive. The inequality holds for the magnitude of the difference.
This can be derived from the main theorem. For example, starting with $a + b > c$, subtract $c$ from both sides: $a+b-c>0$. This doesn't immediately give the difference. Start with $a+b>c$. Subtract $b$ from both sides: $a > c - b$. Also, from $b+c>a$, subtract $c$: $b > a - c$. These inequalities $a > c - b$ and $b > a - c$ combined are equivalent to $|a - c| < b$ when considering the sign of the difference.
Example 1. Is it possible to have a triangle with side lengths 7 cm, 5 cm, and 10 cm?
Answer:
Given: Potential side lengths $a=7$ cm, $b=5$ cm, $c=10$ cm.
To Find: Whether a triangle can be formed with these side lengths.
Solution:
According to the Triangle Inequality Theorem, the sum of the lengths of any two sides must be greater than the length of the third side. We need to check all three inequalities:
- Check if $a + b > c$: $7 \text{ cm} + 5 \text{ cm} = 12 \text{ cm}$. Is $12 \text{ cm} > 10 \text{ cm}$? Yes, this condition is satisfied.
- Check if $a + c > b$: $7 \text{ cm} + 10 \text{ cm} = 17 \text{ cm}$. Is $17 \text{ cm} > 5 \text{ cm}$? Yes, this condition is satisfied.
- Check if $b + c > a$: $5 \text{ cm} + 10 \text{ cm} = 15 \text{ cm}$. Is $15 \text{ cm} > 7 \text{ cm}$? Yes, this condition is satisfied.
Since all three conditions of the Triangle Inequality Theorem are satisfied, it is possible to form a triangle with side lengths 7 cm, 5 cm, and 10 cm.
Example 2. Is it possible to have a triangle with side lengths 3 cm, 4 cm, and 8 cm?
Answer:
Given: Potential side lengths $a=3$ cm, $b=4$ cm, $c=8$ cm.
To Find: Whether a triangle can be formed with these side lengths.
Solution:
We apply the Triangle Inequality Theorem:
- Check if $a + b > c$: $3 \text{ cm} + 4 \text{ cm} = 7 \text{ cm}$. Is $7 \text{ cm} > 8 \text{ cm}$? No, this condition is not satisfied.
Since the sum of two sides (3 cm and 4 cm) is not greater than the third side (8 cm), it is not possible to form a triangle with side lengths 3 cm, 4 cm, and 8 cm.
Note: Once one inequality is found to be false, you don't need to check the others, as the condition for forming a triangle requires all three inequalities to be true.
Triangle Properties: Angles and Sides - Inequalities in a Triangle (Side-Angle Relationships)
We have discussed the exact relationships between the angles of a triangle (sum is $180^\circ$) and the conditions for side lengths to form a triangle (Triangle Inequality Theorem). Now we explore the relationship between the relative sizes of angles and the relative lengths of the sides opposite them. These relationships are expressed as inequalities.
Angle-Side Relationship (Theorem 1)
This theorem relates the lengths of two unequal sides to the measures of the angles opposite them.
Theorem Statement: If two sides of a triangle are unequal, then the angle opposite the longer side is larger (greater) than the angle opposite the shorter side.
Consider $\triangle \text{ABC}$. If $\text{AC} > \text{AB}$, then the angle opposite AC ($\angle \text{B}$) is greater than the angle opposite AB ($\angle \text{C}$). That is, $\text{m}\angle \text{B} > \text{m}\angle \text{C}$.
Proof of the Angle-Side Relationship Theorem
Given: A triangle ABC where the length of side AC is greater than the length of side AB ($\text{AC} > \text{AB}$).
To Prove: $\text{m}\angle \text{ABC} > \text{m}\angle \text{ACB}$.
Construction: Since $\text{AC} > \text{AB}$, we can find a point D on the side AC such that the length of AD is equal to the length of AB ($\text{AD} = \text{AB}$). Join point B to point D to form the line segment BD.
Proof:
Consider the triangle $\triangle \text{ABD}$ formed by the construction.
1. AD = AB
(By Construction)
Since $\triangle \text{ABD}$ has two equal sides, it is an isosceles triangle. By the Isosceles Triangle Theorem, the angles opposite the equal sides are equal.
2. $\angle \text{ABD} = \angle \text{ADB}$
(Angles opposite equal sides in $\triangle \text{ABD}$)
Now, consider $\triangle \text{BDC}$. The angle $\angle \text{ADB}$ is formed by extending side CD of $\triangle \text{BDC}$. Thus, $\angle \text{ADB}$ is an exterior angle for $\triangle \text{BDC}$ at vertex D.
3. $\angle \text{ADB}$ is an exterior angle of $\triangle \text{BDC}$.
(Definition of Exterior Angle)
By the Exterior Angle Property of a triangle, an exterior angle is greater than either of its interior opposite angles.
4. $\angle \text{ADB} > \angle \text{DCB}$
(Exterior Angle Property applied to $\triangle \text{BDC}$)
Note that $\angle \text{DCB}$ is the same angle as $\angle \text{ACB}$.
5. $\angle \text{ADB} > \angle \text{ACB}$
(Substituting notation)
From statement 2 ($\angle \text{ABD} = \angle \text{ADB}$) and statement 5 ($\angle \text{ADB} > \angle \text{ACB}$), by the transitive property of inequality:
6. $\angle \text{ABD} > \angle \text{ACB}$
Now, observe the angle $\angle \text{ABC}$. Point D lies on the side AC, so ray BD is between rays BA and BC. Therefore, $\angle \text{ABC}$ is the sum of $\angle \text{ABD}$ and $\angle \text{DBC}$.
7. $\angle \text{ABC} = \angle \text{ABD} + \angle \text{DBC}$
(Angle Addition Postulate)
Since $\angle \text{DBC}$ is a positive angle measure, from statement 7:
8. $\angle \text{ABC} > \angle \text{ABD}$
(Whole is greater than the part - Common Notion 5)
Finally, combine inequality 8 ($\angle \text{ABC} > \angle \text{ABD}$) and inequality 6 ($\angle \text{ABD} > \angle \text{ACB}$). By the transitive property of inequality:
$\mathbf{\angle \text{ABC} > \angle \text{ACB}}$
Thus, the angle opposite the side AC ($\angle \text{ABC}$) is larger than the angle opposite the side AB ($\angle \text{ACB}$).
Hence Proved.
Side-Angle Relationship (Converse Theorem 2)
This theorem is the converse of the previous one and relates the measures of two unequal angles to the lengths of the sides opposite them.
Theorem Statement: In any triangle, the side opposite to the larger (greater) angle is longer than the side opposite to the smaller angle.
Consider $\triangle \text{ABC}$. If $\text{m}\angle \text{B} > \text{m}\angle \text{C}$, then the side opposite $\angle \text{B}$ (AC) is longer than the side opposite $\angle \text{C}$ (AB). That is, $\text{AC} > \text{AB}$.
Proof of the Side-Angle Relationship Theorem (Converse)
This theorem is typically proven using the method of contradiction (also known as indirect proof).
Given: A triangle ABC where the measure of angle B is greater than the measure of angle C ($\text{m}\angle \text{B} > \text{m}\angle \text{C}$).
To Prove: The length of side AC is greater than the length of side AB ($\text{AC} > \text{AB}$).
Proof:
We will use proof by contradiction. Assume that the conclusion $\text{AC} > \text{AB}$ is false. If $\text{AC} > \text{AB}$ is false, then there are only two other possibilities for the relationship between AC and AB:
Case 1: AC = AB.
Assume AC = AB.
(Assumption for contradiction)
If AC = AB, then $\triangle \text{ABC}$ is an isosceles triangle.
(Definition of Isosceles Triangle)
By the Isosceles Triangle Theorem, the angles opposite these equal sides must be equal.
$\angle \text{B} = \angle \text{C}$
(Angles opposite equal sides)
But this statement ($\angle \text{B} = \angle \text{C}$) directly contradicts the given information that $\angle \text{B} > \angle \text{C}$. Therefore, our assumption that $\text{AC} = \text{AB}$ must be false.
Case 2: AC < AB.
Assume AC < AB.
(Assumption for contradiction)
If $\text{AC} < \text{AB}$, then by the previous theorem (If two sides of a triangle are unequal, the angle opposite to the longer side is larger), the angle opposite the longer side AB ($\angle \text{C}$) must be larger than the angle opposite the shorter side AC ($\angle \text{B}$).
$\angle \text{C} > \angle \text{B}$
(Angle opposite longer side is larger)
But this statement ($\angle \text{C} > \angle \text{B}$) also directly contradicts the given information that $\angle \text{B} > \angle \text{C}$. Therefore, our assumption that $\text{AC} < \text{AB}$ must be false.
Since both assumptions (AC = AB and AC < AB) lead to a contradiction with the given information, the only remaining possibility, which is the negation of our initial assumption, must be true.
The negation of "AC is not greater than AB" is "AC is greater than AB".
Therefore, $\mathbf{\text{AC} > \text{AB}}$.
Thus, the side opposite the larger angle ($\angle \text{B}$) is longer than the side opposite the smaller angle ($\angle \text{C}$).
Hence Proved.
Summary of Side-Angle Inequality Relationships
In any triangle:
- The longest side is opposite the largest angle.
- The shortest side is opposite the smallest angle.
- If two sides are equal, the angles opposite them are equal (Isosceles Triangle Theorem).
- If two angles are equal, the sides opposite them are equal (Converse of Isosceles Triangle Theorem).
These relationships mean there is a direct correspondence between the ordering of angle measures and the ordering of the lengths of the sides opposite those angles.
Example 1. In $\triangle \text{XYZ}$, $\angle \text{X} = 70^\circ$ and $\angle \text{Y} = 50^\circ$. Which side of the triangle is the longest?
Answer:
Given: In $\triangle \text{XYZ}$, $\text{m}\angle \text{X} = 70^\circ$ and $\text{m}\angle \text{Y} = 50^\circ$.
To Find: The longest side of $\triangle \text{XYZ}$.
Solution:
To determine the longest side, we first need to know the measures of all three interior angles so that we can identify the largest angle. We can find the measure of the third angle, $\angle \text{Z}$, using the Angle Sum Property of a triangle.
$\text{m}\angle \text{X} + \text{m}\angle \text{Y} + \text{m}\angle \text{Z} = 180^\circ$
(Angle Sum Property)
Substitute the given angle measures:
$70^\circ + 50^\circ + \text{m}\angle \text{Z} = 180^\circ$
$120^\circ + \text{m}\angle \text{Z} = 180^\circ$
Subtract $120^\circ$ from both sides:
$\text{m}\angle \text{Z} = 180^\circ - 120^\circ$
$\text{m}\angle \text{Z} = 60^\circ$
Now we have the measures of all three interior angles:
$\text{m}\angle \text{X} = 70^\circ$
$\text{m}\angle \text{Y} = 50^\circ$
$\text{m}\angle \text{Z} = 60^\circ$
Comparing the measures, the largest angle is $\angle \text{X}$, with a measure of $70^\circ$.
According to the theorem "the side opposite to the larger angle is longer", the side opposite the largest angle ($\angle \text{X}$) must be the longest side of the triangle.
The side opposite vertex X is the side connecting vertices Y and Z, which is side YZ.
Therefore, side YZ is the longest side of $\triangle \text{XYZ}$.