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, $latex \overline{BD} \cong \overline{BD}$. Continue reading