Question

Read the proof.

Given: AB ∥ DE

Prove: △ACB ~ △DCE

Triangle A B C is shown. Line D E is drawn inside of the triangle and is parallel to side A B. The line forms triangle D C E.

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.

Answer 1

AA similarity theorem.

Reason: ∠CED ≅ ∠CBA (corresponding angles from AB ∥ DE) and ∠C ≅ ∠C (common/reflexive), so two angles are congruent and by AA the triangles are similar.