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 second triangle pair is △AED and △CEB. Their corresponding vertices are A↔C, E↔E, and D↔B, so CPCTC gives ∠EAD ≅ ∠ECB. Since A, E, and C lie on the same line, these are ∠CAD and ∠ACB; congruent alternate interior angles show that AD is parallel to BC. The matching choice is △AED.

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