Triangle Congruence Theorems

When two triangles are congruent, it means they are exactly the same shape and size. All six corresponding parts (three angles and three sides) are equal. However, you do not need to measure all six parts to prove that two triangles are congruent. You can use one of the five Triangle Congruence Theorems to save time.

1. Side-Side-Side (SSS)

If three sides of one triangle are exactly equal in length to three sides of another triangle, then the two triangles are congruent.

• If $AB = DE$, $BC = EF$, and $AC = DF$, then $\triangle ABC \cong \triangle DEF$.

2. Side-Angle-Side (SAS)

If two sides and the included angle (the angle formed directly between those two sides) of one triangle are equal to the corresponding parts of another triangle, the triangles are congruent.

• If $AB = DE$, $\angle B = \angle E$, and $BC = EF$, then $\triangle ABC \cong \triangle DEF$.

3. Angle-Side-Angle (ASA)

If two angles and the included side (the side strictly between the two angles) of one triangle are equal to the corresponding parts of another triangle, the triangles are congruent.

• If $\angle A = \angle D$, $AB = DE$, and $\angle B = \angle E$, then $\triangle ABC \cong \triangle DEF$.

4. Angle-Angle-Side (AAS)

If two angles and a non-included side of one triangle are equal to the corresponding parts of another triangle, the triangles are congruent.

• If $\angle A = \angle D$, $\angle B = \angle E$, and $BC = EF$, then $\triangle ABC \cong \triangle DEF$.

5. Hypotenuse-Leg (HL)

This theorem applies only to right-angled triangles. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, the triangles are congruent.

Example Problems

Example 1: Determine the congruence theorem Determine which congruence theorem proves $\triangle ABC \cong \triangle DEF$ given $AB = DE$, $\angle A = \angle D$, and $\angle B = \angle E$.

Solution: We are given two corresponding angles ($\angle A$ and $\angle B$) and the side directly connecting them ($AB$). Because the side is included between the two given angles, the correct theorem is Angle-Side-Angle (ASA).

Example 2: Prove $\triangle ABC \cong \triangle CDA$ in parallelogram $ABCD$

Solution: In parallelogram $ABCD$, draw the diagonal $AC$. We want to prove $\triangle ABC \cong \triangle CDA$.

1. Opposite sides of a parallelogram are equal in length: $AB = CD$ and $BC = DA$.
2. The diagonal is a shared side for both triangles: $AC = CA$ (Reflexive Property).
3. Since all three corresponding sides are equal, we can confidently conclude that $\triangle ABC \cong \triangle CDA$ by the Side-Side-Side (SSS) theorem.

(Note: You could also prove this using SAS or ASA by utilizing alternate interior angles!)