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.

(A) AA similarity

Step-by-step explanation:

Given: AB is parallel to DE.

To prove: △ACB is similar to △DCE.

Proof: It is given that AB is parallel to DE, thus because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines.

Now, from △ACB and △DCE, we have

∠CED≅∠CBA (corresponding angles of transversal CB and are therefore congruent)

and ∠C ≅ ∠C (Reflexive property)

Thus, by AA similarity postulate,

△ACB is similar to △DCE

Hence proved.

Thus, option A is correct.

i hope this is what you were looking for.

see attached graph