Question

Read the proof.

Given: AB ∥ DE

Prove: △ACB ~ △DCE
e 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.

Answer 1

AA similarity theorem.

You have shown two pairs of congruent angles: ∠C ≅ ∠C (reflexive) and ∠CBA ≅ ∠CED (corresponding angles from AB ∥ DE). By AA, △ACB ~ △DCE (with the correspondence A↔D, C↔C, B↔E).