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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
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| Congruence of Figures and Congruent Triangles | Criteria for Congruence of Triangles | Properties related to Triangles |
| Inequalities in a Triangle | ||
Chapter 7 Triangles (Concepts)
This pivotal chapter delves deep into the fundamental concept of Congruence as applied specifically to triangles, and explores the crucial properties and inequalities that arise within these three-sided figures. Building upon introductory concepts, we now engage with these ideas more formally, focusing on rigorous proof and logical deduction. Understanding triangle congruence is not just about recognizing identical shapes; it's about establishing precise conditions that guarantee this identity and leveraging that knowledge to uncover further properties of geometric figures, particularly isosceles triangles, and the inherent relationships governing their sides and angles.
We begin by rigorously revisiting the core idea of congruence. Recall that two geometric figures are deemed congruent if they possess the exact same shape and size – meaning one can be perfectly superimposed onto the other. When applied to triangles, say $\triangle ABC$ and $\triangle PQR$, congruence implies a precise one-to-one correspondence between their vertices such that all corresponding parts are equal. This means:
- Corresponding sides are equal in length: $AB = PQ$, $BC = QR$, $AC = PR$.
- Corresponding angles are equal in measure: $\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$.
- SSS (Side-Side-Side): If three sides of one triangle are equal to the corresponding three sides of another.
- SAS (Side-Angle-Side): If two sides and the included angle of one triangle are equal to the corresponding parts of another.
- ASA (Angle-Side-Angle): If two angles and the included side of one triangle are equal to the corresponding parts of another.
- RHS (Right angle-Hypotenuse-Side): Applicable only to right-angled triangles, requiring equality of the hypotenuse and one corresponding leg.
Mastering the correct application of these criteria, including the precise statement of correspondence (e.g., stating $\triangle ABC \cong \triangle PQR$ establishes the vertex mapping $A \leftrightarrow P, B \leftrightarrow Q, C \leftrightarrow R$), is paramount. Once congruence is proven, the powerful principle of CPCT (Corresponding Parts of Congruent Triangles are equal) allows us to deduce the equality of any remaining corresponding sides or angles.
A significant application of congruence lies in exploring the properties of isosceles triangles (triangles with at least two equal sides). Two fundamental theorems are rigorously proven, typically using congruence:
- Theorem 1: Angles opposite to equal sides of an isosceles triangle are equal. (If $AB = AC$ in $\triangle ABC$, then $\angle C = \angle B$). The proof often involves constructing the bisector of the angle between the equal sides and proving the two resulting smaller triangles congruent using SAS.
- Theorem 2 (Converse): Sides opposite to equal angles of a triangle are equal. (If $\angle B = \angle C$ in $\triangle ABC$, then $AC = AB$). This proof might also involve an angle bisector and the ASA congruence criterion.
The chapter then transitions into the fascinating realm of inequalities within a triangle, establishing rules that govern the relative sizes of sides and angles when they are not equal. Key theorems proven include:
- Angle-Side Inequality 1: If two sides of a triangle are unequal, the angle opposite to the longer side is larger. (In $\triangle ABC$, if $AC > AB$, then $\angle B > \angle C$).
- Angle-Side Inequality 2 (Converse): In any triangle, the side opposite to the larger angle is longer. (In $\triangle ABC$, if $\angle B > \angle C$, then $AC > AB$).
- Triangle Inequality Property: The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. ($AB + BC > AC$, $BC + AC > AB$, $AC + AB > BC$). This fundamental property determines if three given lengths can even form a triangle.
Congruence of Figures and Congruent Triangles
In geometry, we often compare figures to see if they are the same. While "same shape" refers to similarity, "same shape and same size" refers to congruence. The concept of congruence is fundamental in geometry and is used extensively in proving geometric theorems and solving problems.
Congruence of Geometric Figures
Two geometric figures are said to be congruent if they are exact duplicates of each other. This means they have precisely the same shape and the same size. If two figures are congruent, one can be placed directly on top of the other in such a way that they perfectly overlap and coincide completely. This perfect overlapping is called superposition.
The symbol for congruence is '$\cong$', which is a combination of the symbol for similarity ('$\sim$') and the symbol for equality ('$=$'), indicating same shape and same size.
Examples of Congruent Figures in General:
- Two line segments are congruent if and only if they have the same length. $\overline{AB} \cong \overline{CD}$ if length of $\overline{AB}$ = length of $\overline{CD}$.
- Two squares are congruent if and only if they have the same side length. If Square A has side $s_1$ and Square B has side $s_2$, then Square A $\cong$ Square B if $s_1 = s_2$.
- Two circles are congruent if and only if they have the same radius (or diameter). If Circle C1 has radius $r_1$ and Circle C2 has radius $r_2$, then C1 $\cong$ C2 if $r_1 = r_2$.
- Two angles are congruent if and only if they have the same measure. $\angle ABC \cong \angle PQR$ if measure of $\angle ABC$ = measure of $\angle PQR$.
- Objects produced in a factory using the same mould are usually congruent (e.g., biscuits from the same mould, nuts and bolts of the same specification).
In general, if two figures are congruent, then all their corresponding parts are equal. Conversely, if all corresponding parts of two figures are equal, then the figures are congruent.
Congruent Triangles
Congruence is particularly important when studying triangles. Two triangles are congruent if and only if their vertices can be paired up in such a way that the corresponding sides and the corresponding angles are equal.
If $\triangle ABC$ is congruent to $\triangle PQR$, we write it as $\triangle ABC \cong \triangle PQR$. The order of the vertices in the congruence statement is very important, as it indicates the correspondence between the vertices of the two triangles.
When we write $\triangle ABC \cong \triangle PQR$, it implies the following correspondence between the vertices:
- Vertex A corresponds to Vertex P ($A \leftrightarrow P$)
- Vertex B corresponds to Vertex Q ($B \leftrightarrow Q$)
- Vertex C corresponds to Vertex R ($C \leftrightarrow R$)
This one-to-one correspondence between the vertices means that the corresponding sides and angles are equal:
- Corresponding sides are equal: The side connecting corresponding vertices are equal. Based on the correspondence $A \leftrightarrow P, B \leftrightarrow Q, C \leftrightarrow R$:
- Side AB corresponds to side PQ, so $\overline{AB} = \overline{PQ}$ (or length AB = length PQ).
- Side BC corresponds to side QR, so $\overline{BC} = \overline{QR}$ (or length BC = length QR).
- Side CA corresponds to side RP, so $\overline{CA} = \overline{RP}$ (or length CA = length RP).
- Corresponding angles are equal: The angles at corresponding vertices are equal. Based on the correspondence $A \leftrightarrow P, B \leftrightarrow Q, C \leftrightarrow R$:
- Angle at A corresponds to angle at P, so $\angle A = \angle P$.
- Angle at B corresponds to angle at Q, so $\angle B = \angle Q$.
- Angle at C corresponds to angle at R, so $\angle C = \angle R$.
CPCTC (Corresponding Parts of Congruent Triangles)
The fact that all corresponding sides and angles of congruent triangles are equal is a fundamental property and is often abbreviated as CPCTC. This abbreviation stands for Corresponding Parts of Congruent Triangles are Congruent (or Equal, since congruence of segments/angles means equality of their measures/lengths).
Conversely, if the corresponding sides and angles of two triangles are equal, then the two triangles are congruent. For example, if you have $\triangle ABC$ and $\triangle PQR$ and know that $\overline{AB} = \overline{PQ}$, $\overline{BC} = \overline{QR}$, $\overline{CA} = \overline{RP}$, $\angle A = \angle P$, $\angle B = \angle Q$, and $\angle C = \angle R$, then you can conclude that $\triangle ABC \cong \triangle PQR$.
The order of vertices in the congruence statement $\triangle ABC \cong \triangle PQR$ is fixed by the correspondence: A corresponds to P, B to Q, C to R. If the correspondence was different, say $A \leftrightarrow Q, B \leftrightarrow P, C \leftrightarrow R$, and the triangles were congruent under this correspondence, the statement would be $\triangle ABC \cong \triangle QPR$.
Example 1. If $\triangle$XYZ $\cong$ $\triangle$LMN, write the pairs of corresponding sides and corresponding angles.
Answer:
Given: $\triangle$XYZ $\cong$ $\triangle$LMN.
To Find: Pairs of corresponding sides and corresponding angles.
Solution:
The congruence statement $\triangle$XYZ $\cong$ $\triangle$LMN establishes the vertex correspondence directly by the order of the letters:
- X corresponds to L ($X \leftrightarrow L$)
- Y corresponds to M ($Y \leftrightarrow M$)
- Z corresponds to N ($Z \leftrightarrow N$)
Using this correspondence, the pairs of corresponding sides are:
- Side formed by the first two vertices of $\triangle$XYZ (XY) corresponds to the side formed by the first two vertices of $\triangle$LMN (LM). Thus, XY = LM.
- Side formed by the second and third vertices of $\triangle$XYZ (YZ) corresponds to the side formed by the second and third vertices of $\triangle$LMN (MN). Thus, YZ = MN.
- Side formed by the third and first vertices of $\triangle$XYZ (ZX) corresponds to the side formed by the third and first vertices of $\triangle$LMN (NL). Thus, ZX = NL.
The pairs of corresponding angles are:
- Angle at the first vertex of $\triangle$XYZ ($\angle$X) corresponds to the angle at the first vertex of $\triangle$LMN ($\angle$L). Thus, $\angle$X = $\angle$L.
- Angle at the second vertex of $\triangle$XYZ ($\angle$Y) corresponds to the angle at the second vertex of $\triangle$LMN ($\angle$M). Thus, $\angle$Y = $\angle$M.
- Angle at the third vertex of $\triangle$XYZ ($\angle$Z) corresponds to the angle at the third vertex of $\triangle$LMN ($\angle$N). Thus, $\angle$Z = $\angle$N.
All these equalities hold true because they are corresponding parts of congruent triangles (CPCTC).
Criteria for Congruence of Triangles
Two triangles are congruent if they are exact copies of each other. This means all their corresponding sides and angles are equal. While you could check all six corresponding parts (3 sides and 3 angles), it's not necessary. Mathematics provides us with "shortcuts" called congruence criteria or rules. If a pair of triangles satisfies just one of these criteria, it is a guarantee that they are congruent, and therefore all their other corresponding parts will also be equal (a property we call CPCTC - Corresponding Parts of Congruent Triangles are Congruent).
SAS Congruence Rule (Side-Angle-Side)
The SAS (Side-Angle-Side) Congruence Rule states that two triangles are congruent if two sides and the included angle (the angle formed between those two sides) of one triangle are equal to the corresponding two sides and included angle of the other triangle.
If in $\triangle ABC$ and $\triangle PQR$:
- Side $AB = PQ$
- Included Angle $\angle B = \angle Q$
- Side $BC = QR$
Then, we can conclude that $\triangle ABC \cong \triangle PQR$.
SAS as an Axiom
The SAS rule is so fundamental that it is treated as an axiom in geometry. An axiom is a statement that is accepted as true without proof. We accept SAS as a basic truth because its proof relies on the intuitive idea of "superposition" – physically picking up one triangle and placing it perfectly on top of the other, which is not a formal geometric proof.
ASA Congruence Rule (Angle-Side-Angle)
The ASA (Angle-Side-Angle) Congruence Rule states that two triangles are congruent if two angles and the included side (the side connecting the vertices of those two angles) of one triangle are equal to the corresponding two angles and included side of the other triangle.
Theorem 7.1 (ASA Congruence Rule). Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.
Proof:
Given: Two triangles, $\triangle ABC$ and $\triangle PQR$, such that $\angle B = \angle Q$, $\angle C = \angle R$, and $BC = QR$.
To Prove: $\triangle ABC \cong \triangle PQR$.
Proof:
We will compare the side $AB$ with the side $PQ$. There are three possibilities:
Case 1: Let $AB = PQ$.
In this case, we have:
AB = PQ
(Assumption)
$\angle B = \angle Q$
(Given)
BC = QR
(Given)
Therefore, by the SAS Congruence Axiom, $\triangle ABC \cong \triangle PQR$.
Case 2: Assume $AB > PQ$.
If $AB$ is greater than $PQ$, we can take a point D on side AB such that $DB = PQ$.
Now, consider $\triangle DBC$ and $\triangle PQR$. We have:
DB = PQ
(By construction)
$\angle B = \angle Q$
(Given)
BC = QR
(Given)
So, by the SAS Congruence Axiom, $\triangle DBC \cong \triangle PQR$.
Since the triangles are congruent, their corresponding parts must be equal (CPCTC). Therefore:
$\angle DCB = \angle PRQ$
(CPCTC)
But we were given that:
$\angle ACB = \angle PRQ$
(Given)
This means $\angle DCB = \angle ACB$. This is a contradiction unless point D is the same as point A. If D and A are the same point, then $AB = DB$. But we constructed $DB=PQ$, so this implies $AB=PQ$. This contradicts our assumption that $AB > PQ$.
Case 3: Assume $AB < PQ$.
A similar argument (by taking a point on an extended AB) will also lead to a contradiction.
Since Cases 2 and 3 lead to contradictions, the only possibility is Case 1, where $AB = PQ$. Therefore, by SAS, the triangles are congruent.
AAS Congruence Rule (Angle-Angle-Side)
The AAS (Angle-Angle-Side) Congruence Rule states that two triangles are congruent if any two pairs of angles and one pair of corresponding non-included sides are equal.
Proof: The AAS rule can be easily proven using the ASA rule.
Given $\triangle ABC$ and $\triangle PQR$ where $\angle B = \angle Q$, $\angle C = \angle R$, and $AC = PR$. By the Angle Sum Property of a triangle, the third angles must also be equal:
$\angle A = 180^\circ - (\angle B + \angle C)$
(Angle Sum Property)
$\angle P = 180^\circ - (\angle Q + \angle R)$
(Angle Sum Property)
Since $\angle B = \angle Q$ and $\angle C = \angle R$, it follows that $\angle A = \angle P$.
Now, we have:
$\angle A = \angle P$
(Proven above)
AC = PR
(Given)
$\angle C = \angle R$
(Given)
This fits the ASA (Angle-Side-Angle) condition. Therefore, $\triangle ABC \cong \triangle PQR$.
SSS Congruence Rule (Side-Side-Side)
The SSS (Side-Side-Side) Congruence Rule states that two triangles are congruent if all three sides of one triangle are equal in length to the corresponding three sides of the other triangle.
Theorem (SSS Congruence Rule). If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
Proof:
Given: Two triangles, $\triangle ABC$ and $\triangle PQR$, such that $AB = PQ$, $BC = QR$, and $AC = PR$.
To Prove: $\triangle ABC \cong \triangle PQR$.
Construction: Draw a line segment from C to a point S on the opposite side of BC from A, such that $\angle QPR = \angle SCB$ and $BS = PQ$. Join AS.
Proof:
In $\triangle SBC$ and $\triangle PQR$, we have:
BC = QR
(Given)
$\angle SBC = \angle PQR$
(By Construction)
BS = PQ
(By Construction)
So, by SAS Congruence, $\triangle SBC \cong \triangle PQR$. By CPCTC, this means:
SC = PR
(CPCTC)
$\angle BSC = \angle QPR$
(CPCTC)
We are given that $AC = PR$. Since we proved $SC = PR$, it follows that $AC = SC$. Similarly, since $AB = PQ$ (given) and $BS = PQ$ (construction), it follows that $AB = BS$.
Now, join A to S. Consider $\triangle ABS$ and $\triangle ACS$.
In $\triangle ABS$, $AB = BS$, making it an isosceles triangle. Therefore:
$\angle BAS = \angle BSA$
(Angles opposite equal sides) ... (i)
In $\triangle ACS$, $AC = SC$, making it an isosceles triangle. Therefore:
$\angle CAS = \angle CSA$
(Angles opposite equal sides) ... (ii)
Adding equations (i) and (ii):
$\angle BAS + \angle CAS = \angle BSA + \angle CSA$
(Adding (i) and (ii))
This simplifies to: $\angle BAC = \angle BSC$. But we already know $\angle BSC = \angle QPR$. Thus, $\angle BAC = \angle QPR$.
Now, in our original triangles $\triangle ABC$ and $\triangle PQR$, we have:
AB = PQ
(Given)
$\angle BAC = \angle QPR$
(Proven above)
AC = PR
(Given)
This satisfies the SAS Congruence Axiom. Therefore, $\triangle ABC \cong \triangle PQR$.
RHS Congruence Rule (Right Angle-Hypotenuse-Side)
The RHS (Right Angle-Hypotenuse-Side) Congruence Rule is a special case for right-angled triangles.
Theorem (RHS Congruence Rule). If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.
Proof:
Given: Two right-angled triangles, $\triangle ABC$ and $\triangle PQR$, where $\angle B = \angle Q = 90^\circ$, Hypotenuse $AC = PR$, and Side $AB = PQ$.
To Prove: $\triangle ABC \cong \triangle PQR$.
Proof:
In the right-angled $\triangle ABC$, by Pythagoras' theorem:
$BC^2 = AC^2 - AB^2$
[Pythagoras' Theorem] ... (i)
In the right-angled $\triangle PQR$, by Pythagoras' theorem:
$QR^2 = PR^2 - PQ^2$
[Pythagoras' Theorem] ... (ii)
We are given that $AC = PR$ and $AB = PQ$. Therefore, $AC^2 = PR^2$ and $AB^2 = PQ^2$.
Substituting these into equation (i), we get:
$BC^2 = PR^2 - PQ^2$
[Since AC=PR and AB=PQ] ... (iii)
From equations (ii) and (iii), we have:
$BC^2 = QR^2$
(Both are equal to $PR^2 - PQ^2$)
This implies $BC = QR$.
Now, in $\triangle ABC$ and $\triangle PQR$, we have:
AB = PQ
(Given)
BC = QR
(Proven above)
AC = PR
(Given)
Therefore, by the SSS Congruence Rule, $\triangle ABC \cong \triangle PQR$.
Application Example
Example 1. In quadrilateral ACBD, AC = AD and AB bisects $\angle A$. Show that $\triangle ABC \cong \triangle ABD$. What can you say about BC and BD?
Answer:
Given: Quadrilateral ACBD with AC = AD and AB bisects $\angle CAD$.
To Prove: (i) $\triangle ABC \cong \triangle ABD$ and (ii) State the relationship between BC and BD.
Proof:
Consider the two triangles $\triangle ABC$ and $\triangle ABD$.
AC = AD
(Side - Given)
$\angle CAB = \angle DAB$
(Angle - Since AB bisects $\angle CAD$)
AB = AB
(Side - Common to both triangles)
The angle $\angle CAB$ is included between sides AC and AB. The angle $\angle DAB$ is included between sides AD and AB. Therefore, by the SAS (Side-Angle-Side) congruence rule:
(i) $\triangle ABC \cong \triangle ABD$.
Now, for the second part, since the two triangles are congruent, their corresponding parts must be equal (CPCTC).
The side corresponding to BC in $\triangle ABC$ is BD in $\triangle ABD$.
BC = BD
(By CPCTC)
(ii) Therefore, BC is equal to BD.
Example 2. Line $l$ is the bisector of an angle $\angle A$ and B is any point on $l$. BP and BQ are perpendiculars from B to the arms of $\angle A$. Show that (i) $\triangle APB \cong \triangle AQB$ and (ii) BP = BQ, i.e., B is equidistant from the arms of $\angle A$.
Answer:
Given: Line $l$ bisects $\angle A$. B is on $l$. $BP \perp AP$ and $BQ \perp AQ$.
To Prove: (i) $\triangle APB \cong \triangle AQB$ and (ii) BP = BQ.
Proof:
Consider the two triangles $\triangle APB$ and $\triangle AQB$.
$\angle PAB = \angle QAB$
(Angle - Given that line $l$ bisects $\angle A$)
$\angle APB = \angle AQB$
(Angle - Both are $90^\circ$ as BP and BQ are perpendiculars)
AB = AB
(Side - Common to both triangles)
We have two angles and a non-included side (the common side AB is the hypotenuse) equal in both triangles. Therefore, by the AAS (Angle-Angle-Side) congruence rule:
(i) $\triangle APB \cong \triangle AQB$.
For the second part, since the triangles are congruent, we use CPCTC.
The side BP in $\triangle APB$ corresponds to the side BQ in $\triangle AQB$.
BP = BQ
(By CPCTC)
(ii) Therefore, BP = BQ, which means point B is equidistant from the arms of $\angle A$.
Example 3. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC, and (ii) AD bisects $\angle A$.
Answer:
Given: An isosceles $\triangle ABC$ with $AB=AC$. AD is an altitude to BC, so $AD \perp BC$.
To Prove: (i) BD = CD and (ii) $\angle BAD = \angle CAD$.
Proof:
Since AD is an altitude, it forms two right-angled triangles: $\triangle ADB$ and $\triangle ADC$.
Consider these two right triangles.
$\angle ADB = \angle ADC$
(Right Angle - Both are $90^\circ$ as AD is an altitude)
AB = AC
(Hypotenuse - Given)
AD = AD
(Side - Common to both triangles)
We have a right angle, the hypotenuse, and one other side equal in both triangles. Therefore, by the RHS (Right Angle-Hypotenuse-Side) congruence rule:
$\triangle ADB \cong \triangle ADC$.
Now, we use CPCTC to prove the required results.
(i) The side BD in $\triangle ADB$ corresponds to the side CD in $\triangle ADC$. Therefore:
BD = CD
(By CPCTC)
This shows that AD bisects BC.
(ii) The angle $\angle BAD$ in $\triangle ADB$ corresponds to the angle $\angle CAD$ in $\triangle ADC$. Therefore:
$\angle BAD = \angle CAD$
(By CPCTC)
This shows that AD bisects $\angle A$.
Properties of Triangles
Congruence criteria are powerful tools because they allow us to prove that two triangles are exact copies of each other. Once we establish congruence, we can conclude that all pairs of corresponding sides and angles are equal (CPCTC). We can use these congruence rules to prove important properties about specific types of triangles, such as isosceles and equilateral triangles.
Properties of an Isosceles Triangle
An isosceles triangle is defined as a triangle that has two sides of equal length. The angle formed by the two equal sides is called the vertex angle. The side opposite the vertex angle is called the base. The two angles opposite the equal sides are called the base angles.
In $\triangle ABC$, if $AB = AC$, then $BC$ is the base, and $\angle B$ and $\angle C$ are the base angles.
Theorem 7.2: Angles Opposite Equal Sides
Theorem 7.2. Angles opposite to equal sides of an isosceles triangle are equal.
Proof:
Given:
$\triangle ABC$ is an isosceles triangle in which $AB = AC$.
To Prove:
$\angle B = \angle C$.
Construction:
Draw the bisector of the vertex angle $\angle A$. Let it intersect the base BC at point D.
Proof:
Consider the two triangles formed by the angle bisector: $\triangle ABD$ and $\triangle ACD$.
AB = AC
(Side - Given)
$\angle BAD = \angle CAD$
(Angle - By construction)
AD = AD
(Side - Common)
The angle is included between the two sides. By the SAS (Side-Angle-Side) congruence rule:
$\triangle ABD \cong \triangle ACD$
Since the two triangles are congruent, their corresponding parts are equal (CPCTC).
The angle $\angle B$ in $\triangle ABD$ corresponds to the angle $\angle C$ in $\triangle ACD$.
$\angle B = \angle C$
(By CPCTC)
Thus, the angles opposite the equal sides of an isosceles triangle are equal.
Theorem 7.3: Sides Opposite Equal Angles
Theorem 7.3. The sides opposite to equal angles of a triangle are equal.
Proof:
Given:
$\triangle ABC$ with $\angle B = \angle C$.
To Prove:
$AB = AC$.
Construction:
Draw the angle bisector of $\angle A$, intersecting BC at D.
Proof:
Consider $\triangle ABD$ and $\triangle ACD$.
$\angle B = \angle C$
(Angle - Given)
$\angle BAD = \angle CAD$
(Angle - By construction)
AD = AD
(Side - Common)
We have two angles and a non-included side equal. By the AAS (Angle-Angle-Side) congruence rule:
$\triangle ABD \cong \triangle ACD$
Since the two triangles are congruent, their corresponding parts are equal (CPCTC).
The side AB corresponds to the side AC.
AB = AC
(By CPCTC)
Thus, the sides opposite the equal angles of a triangle are equal.
Theorem 7.4: Altitude from the Vertex of an Isosceles Triangle
Theorem 7.4. The altitude from the vertex of an isosceles triangle to the base bisects the base and also bisects the vertex angle.
Proof:
Given:
$\triangle ABC$ is an isosceles triangle with $AB = AC$. AD is the altitude from vertex A to the base BC, so $AD \perp BC$.
To Prove:
(i) AD bisects BC (i.e., $BD = CD$).
(ii) AD bisects $\angle A$ (i.e., $\angle BAD = \angle CAD$).
Proof:
Consider the two right-angled triangles $\triangle ADB$ and $\triangle ADC$.
$\angle ADB = \angle ADC$
(Right Angle - Both are $90^\circ$ as AD is an altitude)
AB = AC
(Hypotenuse - Given)
AD = AD
(Side - Common)
By the RHS (Right Angle-Hypotenuse-Side) congruence rule:
$\triangle ADB \cong \triangle ADC$
Since the two triangles are congruent, their corresponding parts are equal (CPCTC).
(i) Proving AD bisects BC:
Corresponding sides BD and CD are equal:
BD = CD
(By CPCTC)
This means that D is the midpoint of BC, so AD bisects the base BC.
(ii) Proving AD bisects $\angle A$:
Corresponding angles $\angle BAD$ and $\angle CAD$ are equal:
$\angle BAD = \angle CAD$
(By CPCTC)
This means that AD bisects the vertex angle $\angle A$.
Properties of an Equilateral Triangle
An equilateral triangle is a triangle where all three sides are equal. Since any pair of sides is equal, an equilateral triangle is also an isosceles triangle in three different ways.
Property: Each angle of an equilateral triangle is $60^\circ$.
Proof:
In $\triangle ABC$, let $AB = BC = AC$.
Since $AB = AC$, by Theorem 7.2, the angles opposite them are equal: $\angle C = \angle B$.
Since $BC = AC$, by Theorem 7.2, the angles opposite them are equal: $\angle A = \angle B$.
Therefore, $\angle A = \angle B = \angle C$.
By the Angle Sum Property of a triangle:
$\angle A + \angle B + \angle C = 180^\circ$
$\angle A + \angle A + \angle A = 180^\circ$
$3\angle A = 180^\circ \implies \angle A = 60^\circ$
So, $\angle A = \angle B = \angle C = 60^\circ$.
Application Examples
Example 1. In $\triangle ABC$, AD is the perpendicular bisector of BC. Show that $\triangle ABC$ is an isosceles triangle in which AB = AC.
Answer:
Given:
In $\triangle ABC$, AD is the perpendicular bisector of BC. This means that $AD \perp BC$ (so $\angle ADB = \angle ADC = 90^\circ$) and D is the midpoint of BC (so $BD = CD$).
To Prove:
$\triangle ABC$ is an isosceles triangle, i.e., $AB = AC$.
Proof:
Consider the two triangles $\triangle ADB$ and $\triangle ADC$.
BD = CD
(Side - Given, as AD bisects BC)
$\angle ADB = \angle ADC$
(Angle - Both are $90^\circ$, given)
AD = AD
(Side - Common)
The angle is included between the two sides. By the SAS congruence rule:
$\triangle ADB \cong \triangle ADC$
Since the two triangles are congruent, their corresponding parts are equal (CPCTC).
The sides AB and AC are corresponding sides.
AB = AC
(By CPCTC)
Since two sides of $\triangle ABC$ are equal, $\triangle ABC$ is an isosceles triangle.
Example 2. E and F are respectively the mid-points of equal sides AB and AC of $\triangle ABC$. Show that BF = CE.
Answer:
Given:
In $\triangle ABC$, we are given that $AB=AC$. E is the mid-point of AB, and F is the mid-point of AC.
To Prove:
$BF = CE$.
Proof:
To prove that the line segments BF and CE are equal, we will prove the congruence of two triangles that contain these segments as sides. Let's consider $\triangle ABF$ and $\triangle ACE$.
In $\triangle ABF$ and $\triangle ACE$:
AB = AC
(Side - Given)
$\angle A = \angle A$
(Angle - Common to both triangles)
Since E is the mid-point of AB and F is the mid-point of AC, we have:
AE = $\frac{1}{2}$AB
AF = $\frac{1}{2}$AC
As $AB = AC$, their halves must also be equal. Therefore:
AE = AF
(Side - Halves of equals are equal)
In $\triangle ACE$, the angle $\angle A$ is included between the sides AC and AE. In $\triangle ABF$, the angle $\angle A$ is included between the sides AB and AF. Thus, by the SAS (Side-Angle-Side) congruence rule:
$\triangle ACE \cong \triangle ABF$
Since the triangles are congruent, their corresponding parts are equal (CPCTC).
The side CE in $\triangle ACE$ corresponds to the side BF in $\triangle ABF$.
CE = BF
(By CPCTC)
Hence Proved.
Alternate Solution:
We can also prove this using a different pair of triangles: $\triangle BCF$ and $\triangle CBE$.
In $\triangle BCF$ and $\triangle CBE$:
Since AB = AC is given, the angles opposite to these sides are equal.
$\angle ABC = \angle ACB$
(Angle - Angles opposite equal sides)
Since E is the mid-point of AB and F is the mid-point of AC, and $AB=AC$:
BE = $\frac{1}{2}$AB and CF = $\frac{1}{2}$AC
BE = CF
(Side - Halves of equals are equal)
BC = BC
(Side - Common base)
In $\triangle CBE$, the angle $\angle ABC$ is included between sides BE and BC. In $\triangle BCF$, the angle $\angle ACB$ is included between sides CF and BC. Thus, by the SAS congruence rule:
$\triangle CBE \cong \triangle BCF$
Since the triangles are congruent, their corresponding parts are equal (CPCTC).
The side CE in $\triangle CBE$ corresponds to the side BF in $\triangle BCF$.
CE = BF
(By CPCTC)
Hence Proved.
Inequalities in a Triangle
In previous sections, we focused on conditions for congruence (equality of triangles) and properties of isosceles and equilateral triangles (equality of sides and angles). However, not all triangles have equal sides or angles. In this section, we explore the relationships between the sides and angles of a triangle when they are unequal, leading to important inequalities.
Relationships Between Sides and Angles (Unequal Cases)
There is a direct relationship between the length of a side and the measure of the angle opposite to it in any triangle. The longer the side, the larger the angle opposite to it, and vice versa.
Theorem 7.6: Angle Opposite Longer Side is Larger
This theorem formalises the relationship between unequal sides and the angles opposite them.
Theorem 7.6. If two sides of a triangle are unequal, then the angle opposite to the longer side is larger (greater) than the angle opposite to the shorter side.
Proof:
Given:
$\triangle$ABC with AC $>$ AB.
To Prove:
$\angle$ABC $>$ $\angle$ACB (i.e., $\angle B > \angle C$). The angle opposite AC is $\angle B$, and the angle opposite AB is $\angle C$.
Construction:
On side AC, which is longer than AB, take a point D such that AD = AB. Since AC $>$ AB, point D will lie between A and C. Join BD.
Proof:
Consider $\triangle$ABD. By construction, AD = AB.
Therefore, $\triangle$ABD is an isosceles triangle.
By Theorem 7.2 (Angles opposite equal sides of an isosceles triangle are equal), the angles opposite the equal sides AB and AD are equal:
$\angle$ABD = $\angle$ADB
[Angles opposite equal sides]
$\angle$ABD = $\angle$ADB
... (1)
Now, consider $\angle$ADB. This angle is an exterior angle of $\triangle$BDC (formed by extending side BD of $\triangle$BDC). (Alternatively, think of extending side CD to A; then $\angle ADB$ is exterior to $\triangle BDC$).
By the Exterior Angle Theorem (Theorem 6.8), an exterior angle of a triangle is equal to the sum of its two interior opposite angles.
For $\triangle$BDC, the exterior angle $\angle$ADB is opposite to interior angles $\angle$DBC (or $\angle$CBD) and $\angle$DCB (or $\angle$C$).
$\angle$ADB = $\angle$DBC + $\angle$DCB
[Exterior Angle Theorem]
$\angle$ADB = $\angle$DBC + $\angle$C
... (2)
From equation (2), since $\angle$DBC is a positive angle measure, $\angle$ADB must be greater than $\angle$C (the whole is greater than a part). (Formally, $\angle$ADB - $\angle$C = $\angle$DBC > 0).
$\angle$ADB $>$ $\angle$C
... (3)
From (1) and (3), since $\angle$ABD = $\angle$ADB and $\angle$ADB $>$ $\angle$C, by transitivity of inequality:
$\angle$ABD $>$ $\angle$C
... (4)
Now, consider $\angle$ABC. It is the entire angle at vertex B in $\triangle$ABC. From the figure, $\angle$ABC is composed of $\angle$ABD and $\angle$DBC.
$\angle$ABC = $\angle$ABD + $\angle$DBC
Since $\angle$DBC is a positive angle measure, $\angle$ABC is greater than $\angle$ABD. (Formally, $\angle$ABC - $\angle$ABD = $\angle$DBC > 0).
$\angle$ABC $>$ $\angle$ABD
... (5)
Combining the inequalities (4) and (5), we have $\angle$ABC $>$ $\angle$ABD and $\angle$ABD $>$ $\angle$C. By transitivity, $\angle$ABC $>$ $\angle$C.
$\angle$ABC $>$ $\angle$ACB
Thus, the angle opposite to the longer side (AC) is larger ($\angle B$) than the angle opposite to the shorter side (AB, which is less than AC). The theorem is proven.
Theorem 7.7: Side Opposite Larger Angle is Longer (Converse)
This theorem is the converse of Theorem 7.6 and is also true. It states that in any triangle, the side opposite to the larger angle is longer than the side opposite to the smaller angle.
Theorem 7.7. In any triangle, the side opposite to the larger (greater) angle is longer.
Proof:
Given:
$\triangle$ABC with $\angle B > \angle C$.
To Prove:
AC $>$ AB. (The side opposite $\angle B$ is AC, and the side opposite $\angle C$ is AB).
Proof:
We will prove this theorem using the method of contradiction. For any two sides of a triangle, there are three possible relationships between their lengths: less than, equal to, or greater than.
Consider the sides AC and AB. The three possibilities are:
(i) AC = AB
(ii) AC $<$ AB
(iii) AC $>$ AB
Let's examine cases (i) and (ii) and show that they contradict the given information $\angle B > \angle C$.
Case (i): Assume AC = AB.
If AC = AB, then $\triangle$ABC is an isosceles triangle. By Theorem 7.2 (Angles opposite equal sides are equal), the angles opposite these sides must be equal: $\angle B = \angle C$.
However, this conclusion ($\angle B = \angle C$) contradicts the given information that $\angle B > \angle C$. Therefore, the assumption AC = AB must be false.
Case (ii): Assume AC $<$ AB.
If AC $<$ AB, then applying Theorem 7.6 (Angle opposite the longer side is larger), the angle opposite the longer side AB ($\angle C$) must be greater than the angle opposite the shorter side AC ($\angle B$). So, $\angle C > \angle B$.
However, this conclusion ($\angle C > \angle B$) contradicts the given information that $\angle B > \angle C$. Therefore, the assumption AC $<$ AB must be false.
Case (iii):
Since cases (i) and (ii) have been shown to be false, the only remaining possibility among the three is AC $>$ AB.
Therefore, AC $>$ AB.
Thus, in any triangle, the side opposite to the larger angle is longer. The theorem is proven.
Triangle Inequality Property
This property relates the sum of the lengths of any two sides of a triangle to the length of the third side. It is a fundamental inequality that must hold true for any set of three lengths to form a triangle.
Theorem 7.8: Sum of Two Sides is Greater Than the Third Side
Theorem 7.8. The sum of any two sides of a triangle is greater than the third side.
Proof (Idea using construction and angle-side relationship):
Given:
A triangle ABC.
To Prove:
We need to prove the following three inequalities:
(i) AB + AC $>$ BC
(ii) AB + BC $>$ AC
(iii) BC + AC $>$ AB
Since the proof for all three is similar, we will prove the first one, AB + AC $>$ BC.
Construction:
Extend side BA to a point D such that the length of the extended segment AD is equal to the length of side AC (AD = AC). Join point D to point C.
Proof:
Consider $\triangle$ADC. By construction, AD = AC.
Therefore, $\triangle$ADC is an isosceles triangle.
By Theorem 7.2 (Angles opposite equal sides are equal), the angles opposite the equal sides AD and AC are equal:
$\angle$ADC = $\angle$ACD
$\angle$ADC = $\angle$ACD
... (1)
Now consider the angle $\angle$BCD. From the figure, $\angle$BCD is the entire angle at vertex C in the larger $\triangle$BDC. It is composed of $\angle$BCA (or $\angle C$ of $\triangle$ABC) and $\angle$ACD.
$\angle$BCD = $\angle$BCA + $\angle$ACD
Since $\angle$BCA is a positive angle measure (it's an angle of a triangle), $\angle$BCD is greater than $\angle$ACD. (The whole angle $\angle$BCD is greater than its part $\angle$ACD).
$\angle$BCD $>$ $\angle$ACD
... (2)
From (1) and (2), since $\angle$ADC = $\angle$ACD and $\angle$BCD $>$ $\angle$ACD, by transitivity of inequality:
$\angle$BCD $>$ $\angle$ADC
The angle $\angle$ADC is the same as the angle $\angle$BDC (since D lies on the line segment BD). So, $\angle$BCD $>$ $\angle$BDC.
$\angle$BCD $>$ $\angle$BDC
... (3)
Now consider the larger triangle $\triangle$BDC. We have shown that $\angle$BCD is greater than $\angle$BDC.
By Theorem 7.7 (In any triangle, the side opposite to the larger angle is longer), the side opposite the larger angle $\angle$BCD (which is BD) must be longer than the side opposite the smaller angle $\angle$BDC (which is BC).
BD $>$ BC
... (4)
From the construction, the line segment BD is composed of BA and AD.
BD = BA + AD
Since AD = AC (by construction), substitute this into the equation for BD:
BD = AB + AC
... (5)
Substitute equation (5) into inequality (4):
AB + AC $>$ BC
This proves the first part of the theorem: the sum of sides AB and AC is greater than the third side BC.
Similarly, by extending side AB to E such that BE=BC and joining EC, we can prove AC + BC > AB. By extending side AC to F such that CF=CB and joining FB, we can prove AB + BC > AC. (The methods are analogous, using the same principles of isosceles triangles and the Exterior Angle Theorem).
Thus, the sum of any two sides of a triangle is greater than the third side. The theorem is proven.
Deduction: Difference of Two Sides is Less Than the Third Side
From the Triangle Inequality Property (Theorem 7.8), we can deduce another important relationship. Consider the inequality AB + AC $>$ BC. If we subtract AC from both sides, we get AB $>$ BC - AC. This means the difference between sides BC and AC is less than side AB. Similarly, AB + BC $>$ AC implies AB $>$ AC - BC. Combining these, the difference between any two sides (in absolute value) is less than the third side.
The difference between any two sides of a triangle is less than the third side.
$|a - b| < c$, where a, b, c are the lengths of the sides of a triangle.
Shortest Distance from a Point to a Line
The Triangle Inequality property has implications for the shortest distance between a point and a line.
Theorem 7.9: The Perpendicular is the Shortest Distance
Theorem 7.9. Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
Proof:
Given:
A line $l$ and a point P not lying on the line $l$.
Draw a line segment PM from point P perpendicular to line $l$, where M is the point where the perpendicular meets the line $l$ (PM $\perp l$).
Take any other point Q on the line $l$ (where Q is not M). Draw the line segment PQ.
To Prove:
PM $<$ PQ.
Proof:
Consider $\triangle$PMQ. Since PM is perpendicular to line $l$, $\angle$PMQ is a right angle ($90^\circ$).
$\angle$PMQ = $90^\circ$
In a right-angled triangle, the right angle is the largest angle (since the sum of the other two angles must be $180^\circ - 90^\circ = 90^\circ$, so each of the other two angles is less than $90^\circ$).
In $\triangle$PMQ, $\angle$PMQ is the largest angle.
By the Angle Sum Property of a triangle, $\angle$MPQ + $\angle$PMQ + $\angle$MQP = $180^\circ$.
$\angle$MPQ + $90^\circ$ + $\angle$MQP = $180^\circ$
$\angle$MPQ + $\angle$MQP = $90^\circ$
Since both $\angle$MPQ and $\angle$MQP are parts of this sum and must be positive angles, both $\angle$MPQ $<$ $90^\circ$ and $\angle$MQP $<$ $90^\circ$.
Comparing $\angle$PMQ and $\angle$MQP, we have $\angle$PMQ = $90^\circ$ and $\angle$MQP $<$ $90^\circ$.
$\angle$PMQ $>$ $\angle$MQP
In $\triangle$PMQ, the side opposite to $\angle$PMQ is PQ (the hypotenuse). The side opposite to $\angle$MQP is PM (one of the legs).
By Theorem 7.7 (The side opposite to the larger angle is longer), the side opposite the larger angle ($\angle$PMQ) must be longer than the side opposite the smaller angle ($\angle$MQP$).
PQ $>$ PM
This is equivalent to PM $<$ PQ.
Since Q was any point on the line $l$ other than M, this proves that the perpendicular segment PM is shorter than any other line segment from P to the line $l$.
The length of the perpendicular from a point to a line is defined as the distance of the point from the line, as it is the shortest distance.
Applying Triangle Inequalities
The triangle inequality properties are used to determine if a triangle can be formed with given side lengths or to compare side lengths or angles in various triangle problems.
Example 1. Is it possible to have a triangle with sides of lengths 3 cm, 4 cm, and 6 cm?
Answer:
Given side lengths: $a=3$ cm, $b=4$ cm, $c=6$ cm.
To Determine: If these lengths can form a triangle.
Solution:
According to the Triangle Inequality Property (Theorem 7.8), the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. We must check all three combinations:
1. Check if $a + b > c$:
$3 + 4 > 6$
$7 > 6$
(True)
2. Check if $a + c > b$:
$3 + 6 > 4$
$9 > 4$
(True)
3. Check if $b + c > a$:
$4 + 6 > 3$
$10 > 3$
(True)
Since all three triangle inequalities hold true, it is possible to have a triangle with sides of lengths 3 cm, 4 cm, and 6 cm.