# Sample Proof on Triangle Congruence Part 2

This is the second part on a series of posts on worked proofs in Triangle Congruence. In this post, we prove another problem.

Given: $\overline{AB}$ and $\overline{CD}$ bisect each other at $E$.
Prove: $AD \cong BC$

Proof
A bisector divides a segment into equal parts. Now, since $AB$ and $CD$ bisect each other,
$AE = BE$ and $CE = DE$ by definition of bisector.

We also know that
$\angle AED \cong \angle CEB$ because vertical angles are congruent.

Now, since the angle is between the congruent sides,
$\triangle AED \cong \triangle BEC$ by SAS congruence.

Therefore, $AD \cong BC$ since corresponding sides of congruent triangles are congruent.