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

a. aa similarity theorem. b.sss similarity theorem. c.aas similarity theorem. d.asa similarity theorem.

Given that 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.

From the proof, we have been able to establish two congruencies, that is that angles CED and CBA are congruent and angle C is congruent to angle C.

Therefore, we conclude that △ACB ~ △DCE by the AA similarlity theorem.