Question

read the proof. given: ab // de prove: △acb ~ △dce we are given ab // de. because the lines are parallel and segment cb crosses both lines, we can consider segment cb a transversal of the parallel lines. angles ced and cba are corresponding angles of transversal cb and are therefore congruent, so ∠ced ≅ ∠cba. we can state ∠c ≅ ∠c using the reflexive property. therefore, △acb ~ △dce by the aa similarity theorem. sss similarity theorem. aas similarity theorem. asa similarity theorem.

Explanation:

Two pairs of angles are congruent: ∠CED ≅ ∠CBA and ∠C ≅ ∠C. AA similarity theorem uses two - angle congruence for triangle similarity.

Answer:

A. AA similarity theorem