Question

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

Answer: AAA similarity.

Step-by-step explanation:  CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have

∠CED ≅ ∠CBA,

∠CDE ≅ ∠CAB

and

∠DCE ≅ ∠ACB [same angle]

Hence, by AAA (angle-angle-angle) similarity,

△CED ~ △ABC.

Thus, the correct option is AAA similarity.

Answer 2

Thank you for posting you question here at brainly. Among the choices provided above the answer would be AA similarity theorem. I hope the answer will help you.