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

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

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

**Proof**

It is given that $latex \overline{BE} \cong \overline{DC}$.

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

Notice that $latex \overline{BE}$ is the hypotenuse of $latex \triangle DBE$ and $latex \overline{DC}$ is the hypotenuse of $latex \triangle BDC$. Since both are right triangles

$latex \triangle BDC \cong \triangle DBE$ are congruent by the **Hypotenuse Leg Theorem**.

Next, we prove that $latex \triangle BDC \cong \triangle CAB$.

It is given that $latex \overline{BD} \cong \overline{CA}$.

$latex \angle ACB \cong \angle DBC$ since they are both right angles.

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

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

Now, since $latex \triangle BDC \cong \triangle DBE$, and $latex \triangle BDC \cong CAB$,

$latex \triangle DBE \cong \triangle CAB$ by Transitive Property of Congruence.

This is what we want to prove.

*Reference: UCSMP, Geometry*