# Sample Proof on Triangle Congruence Part 3

Recall that for real numbers a, b, and c are real numbers, if $a = b$ and $b = c$, then $a = c$. This is called Transitive Property of Equality. This is also the same with congruence. If $A$, $B$, and $C$ are polygons, and if $A$ is congruent to $B$, and $B$ is congruent $C$, then $A$ is congruent to $C$. This is called the Transitive Property of Congruence. We will use this to prove the following problem.

Given: $\overline{BE} \cong \overline{DC}$ and $\overline{BD} \cong \overline{CA}$.

Prove: $\triangle DBE \cong \triangle CAB$.

Proof

It is given that $\overline{BE} \cong \overline{DC}$. Now, by reflexive property, that is a segment is congruent to itself, $\overline{BD} \cong \overline{BD}$

Notice that $\overline{BE}$ is the hypotenuse of $\triangle DBE$ and $\overline{DC}$ is the hypotenuse of $\triangle BDC$. Since both are right triangles $\triangle BDC \cong \triangle DBE$ are congruent by the Hypotenuse Leg Theorem.

Next, we prove that $\triangle BDC \cong \triangle CAB$. It is given that $\overline{BD} \cong \overline{CA}$. $\angle ACB \cong \angle DBC$ since they are both right angles.

Now, $\overline{BC} \cong \overline{BC}$ by reflexive property.

Therefore, $\triangle BDC \cong \triangle CAB$ by the SAS Congruence Theorem.

Now, since $\triangle BDC \cong \triangle DBE$, and $\triangle BDC \cong CAB$, $\triangle DBE \cong \triangle CAB$ by Transitive Property of Congruence.

This is what we want to prove.

Reference: UCSMP, Geometry