Question

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.

Answer 1

pls. see attachment. 

Answer 2

Answer:

(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.