MathHub US Geometry Chapter 5

Chapter 5: Congruent Triangles

High School Geometry · Unit 3: Triangles · 3 interactive diagrams · 8 practice problems

Learning Objectives

5.1 Congruent Figures and Corresponding Parts

Two figures are congruent if they have the same shape and size — one can be mapped onto the other by rigid motions (translations, rotations, reflections) only.

Definition: Congruent Triangles

$\triangle ABC \cong \triangle DEF$ means that all six pairs of corresponding parts are congruent:

Order matters! The order of vertices in the congruence statement specifies which parts correspond: first vertex ↔ first vertex, second ↔ second, etc.

Example 5.1 — Reading a Congruence Statement

If $\triangle PQR \cong \triangle XYZ$, list all six pairs of congruent parts.

Sides: $PQ \cong XY$, $QR \cong YZ$, $PR \cong XZ$

Angles: $\angle P \cong \angle X$, $\angle Q \cong \angle Y$, $\angle R \cong \angle Z$

5.2 Triangle Congruence Postulates and Theorems

We don't need to verify all six pairs to prove triangles congruent. Five valid shortcuts (and two invalid ones) are shown below:

SSS
Side-Side-Side: all 3 pairs of sides congruent
SAS
Side-Angle-Side: 2 sides and the included angle
ASA
Angle-Side-Angle: 2 angles and the included side
AAS
Angle-Angle-Side: 2 angles and a non-included side
HL
Hypotenuse-Leg: right triangles only
AAA ✗
Does NOT prove congruence (similar only)
SSA ✗
Does NOT guarantee congruence (ambiguous case)

The Five Congruence Shortcuts

Memory Aid: "SAS means the angle is sandwiched between the two sides." For SSA — remember "a donkey is stubborn" — SSA doesn't work! The angle must be included between the sides for SAS, and the side must be included between the angles for ASA.

Example 5.2 — Identifying the Correct Postulate

Which postulate or theorem proves $\triangle JKL \cong \triangle MNP$?

TRY IT

Right $\triangle ABC$ and $\triangle DEF$ have right angles at $B$ and $E$. If $AC = DF$ (hypotenuses) and $AB = DE$ (legs), which postulate proves congruence?

Show Answer
HL (Hypotenuse-Leg) — both are right triangles, the hypotenuses are equal, and one pair of legs is equal. This is sufficient to prove $\triangle ABC \cong \triangle DEF$.

SSS, SAS, ASA, AAS illustrated — pairs of congruent triangles with color-coded matching parts.

Figure 5.1 — Congruent Triangle Pairs by Different Postulates

5.3 CPCTC: Corresponding Parts of Congruent Triangles Are Congruent

CPCTC

CPCTC stands for "Corresponding Parts of Congruent Triangles Are Congruent." Once you prove two triangles congruent, you can conclude that any pair of corresponding parts are also congruent. CPCTC is used as a reason in the later steps of a proof, after establishing the triangle congruence.

The typical two-column proof structure when using CPCTC:

  1. Establish congruence of specific sides and angles (given or derived)
  2. State the triangle congruence (SSS / SAS / ASA / AAS / HL)
  3. Conclude additional corresponding parts are congruent by CPCTC

Example 5.3 — A Complete Two-Column Proof with CPCTC

Given: $\overline{AB} \cong \overline{CB}$; $B$ is the midpoint of $\overline{AC}$ — wait, let's use a cleaner setup:

Given: $\overline{AB} \cong \overline{CD}$; $\overline{AB} \parallel \overline{CD}$

Prove: $\triangle ABE \cong \triangle DCE$ where $E$ is the intersection of $\overline{AD}$ and $\overline{BC}$

StatementsReasons
$AB \cong CD$Given
$AB \parallel CD$Given
$\angle ABE \cong \angle DCE$Alternate interior angles, $AB \parallel CD$, transversal $\overline{BC}$
$\angle BAE \cong \angle CDE$Alternate interior angles, $AB \parallel CD$, transversal $\overline{AD}$
$\triangle ABE \cong \triangle DCE$AAS
$BE \cong CE$ and $AE \cong DE$CPCTC
TRY IT

In the proof above, what additional conclusion can you draw from CPCTC? (Hint: what does $AE \cong DE$ and $BE \cong CE$ tell you about point $E$?)

Show Answer
$AE \cong DE$ means $E$ is the midpoint of $\overline{AD}$. $BE \cong CE$ means $E$ is the midpoint of $\overline{BC}$. Therefore, the diagonals of quadrilateral $ABCD$ bisect each other — proving $ABCD$ is a parallelogram!

5.4 Isosceles Triangle Theorem

Isosceles Triangle Theorem and Converse

Example 5.4 — Isosceles Triangle

In $\triangle ABC$, $AB = AC = 7$ and $\angle ABC = 52°$. Find $\angle BAC$.

By the Isosceles Triangle Theorem, $\angle ABC = \angle ACB = 52°$.

$\angle BAC = 180° - 52° - 52° = \mathbf{76°}$

Isosceles triangle with vertex angle slider — observe that the base angles remain equal as the shape changes.

Figure 5.2 — Isosceles Triangle: Base Angles Always Equal

SSS vs. SAS comparison — both postulate pairs produce the same unique triangle shape, confirming congruence.

Figure 5.3 — SSS and SAS Produce Unique (Congruent) Triangles

Practice Problems

1

$\triangle ABC \cong \triangle PQR$. If $AB = 3x + 2$ and $PQ = 5x - 6$, find $AB$.

Show Solution
Corresponding sides: $AB \cong PQ$, so $3x + 2 = 5x - 6 \Rightarrow 8 = 2x \Rightarrow x = 4$. Then $AB = 3(4) + 2 = \mathbf{14}$.
2

For each set of given info, identify the congruence postulate or state "not enough":
(a) Two sides and the included angle
(b) Two angles and any side
(c) Two sides and a non-included angle
(d) Three angles only

Show Solution
(a) SAS. (b) AAS (if non-included side) or ASA (if included side). (c) Not enough (SSA — ambiguous case). (d) Not enough (AAA — proves similarity only, not congruence).
3

$\overline{WX}$ bisects $\angle VWY$ and $\overline{WX} \perp \overline{VY}$. Prove $\triangle VWX \cong \triangle YWX$. What postulate applies?

Show Solution
Given: $WX$ bisects $\angle VWY \Rightarrow \angle VWX = \angle YWX$. Given: $WX \perp VY \Rightarrow \angle WXV = \angle WXY = 90°$. Shared: $WX \cong WX$ (reflexive). So we have two angles ($\angle VWX = \angle YWX$, right angles) and the included side $WX$. This is ASA.
4

In isosceles $\triangle KLM$ with $KL = KM$, $\angle L = 3x + 10$ and $\angle M = 5x - 4$. Find all angles.

Show Solution
Base angles equal: $3x + 10 = 5x - 4 \Rightarrow 14 = 2x \Rightarrow x = 7$. So $\angle L = \angle M = 31°$ and $\angle K = 180° - 31° - 31° = \mathbf{118°}$.
5

In the diagram, $\overline{AC}$ is a common side of $\triangle ABC$ and $\triangle ACD$. Given $AB \cong CD$ and $BC \cong AD$, prove the triangles congruent. (Which postulate?)

Show Solution
$AB \cong CD$ (given), $BC \cong AD$ (given), $AC \cong AC$ (reflexive property). Three pairs of sides are congruent, so $\triangle ABC \cong \triangle CDA$ by SSS.
6

$\triangle ABC \cong \triangle DEF$ (proved by ASA). Can you conclude $BC \cong EF$? What about $\angle A \cong \angle D$? State your reasoning.

Show Solution
Yes to both. Since $\triangle ABC \cong \triangle DEF$, by CPCTC, all corresponding parts are congruent. So $BC \cong EF$ (corresponding sides) and $\angle A \cong \angle D$ (corresponding angles), regardless of which shortcut was used to prove congruence.
7

Two-column proof: Given $M$ is the midpoint of $\overline{AB}$ and $M$ is the midpoint of $\overline{CD}$. Prove $\triangle AMC \cong \triangle BMD$.

Show Solution
Statements / Reasons:
1. $M$ is midpoint of $AB$ / Given
2. $AM \cong MB$ / Definition of midpoint
3. $M$ is midpoint of $CD$ / Given
4. $CM \cong MD$ / Definition of midpoint
5. $\angle AMC \cong \angle BMD$ / Vertical angles
6. $\triangle AMC \cong \triangle BMD$ / SAS $\square$
8

In right $\triangle RST$ (right angle at $S$) and right $\triangle UVW$ (right angle at $V$), $RT = UW$ and $RS = UV$. Prove $\triangle RST \cong \triangle UVW$ and find $\angle T$ if $\angle R = 35°$.

Show Solution
Both are right triangles. $RT = UW$ are the hypotenuses, $RS = UV$ are legs. By HL, $\triangle RST \cong \triangle UVW$. Then $\angle T = \angle W$ (CPCTC). By Angle Sum: $\angle T = 180° - 90° - 35° = \mathbf{55°}$, and $\angle W = 55°$.

📋 Chapter Summary

Congruence Postulates and Theorems

SSS (Side-Side-Side)

If three sides of one triangle are congruent to three sides of another, the triangles are congruent.

SAS (Side-Angle-Side)

Two sides and the included angle (between those sides) are congruent. The angle must be between the two sides.

ASA (Angle-Side-Angle)

Two angles and the included side are congruent. The side must be between the two angles.

AAS (Angle-Angle-Side)

Two angles and a non-included side are congruent. Different from ASA — the side is not between the angles.

HL (Hypotenuse-Leg)

For right triangles only: hypotenuse and one leg are congruent. The right angle is already known.

CPCTC

"Corresponding Parts of Congruent Triangles are Congruent" — used after proving triangles congruent to conclude other parts are equal.

Isosceles Triangle Theorem

  1. Base angles of an isosceles triangle are congruent
  2. Converse: if two angles of a triangle are congruent, the sides opposite them are congruent
  3. The angle bisector from the vertex angle of an isosceles triangle is also the perpendicular bisector of the base

📘 Key Terms

CongruentHaving the same shape and size. All corresponding sides and angles are equal.
Included AngleThe angle between two given sides of a triangle.
Included SideThe side between two given angles of a triangle.
CPCTCCorresponding Parts of Congruent Triangles are Congruent — a reason used after establishing congruence.
Isosceles TriangleA triangle with at least two congruent sides. The base angles (opposite the equal sides) are congruent.
HypotenuseThe longest side of a right triangle, opposite the right angle.
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