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

Given: $latex \overline{AB}$ and $latex \overline{CD}$ bisect each other at $latex E$.

Prove: $latex AD \cong BC$

**Proof**

A bisector divides a segment into equal parts. Now, since $latex AB$ and $latex CD$ bisect each other,

$latex AE = BE$ and $latex CE = DE$ by definition of bisector.

We also know that

$latex \angle AED \cong \angle CEB$ because vertical angles are congruent.

Now, since the angle is between the congruent sides,

$latex \triangle AED \cong \triangle BEC$ by SAS congruence.

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