By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are parallel, quadrilateral ABCD is a parallelogram.

The matching triangle is △AED: its corresponding vertices are A↔C, E↔E, and D↔B, so △AED ≅ △CEB by the same congruence argument. Then CPCTC gives ∠EAD ≅ ∠ECB; since E lies on AC, these are ∠CAD and ∠ECB. These are alternate interior angles, so AD is parallel to BC.

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