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

Answer: AA similarity theorem.

Step-by-step explanation:

Given : AB ∥ DE

Prove: ΔACB ≈ ΔDCE

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.

Also ∠C ≅ ∠C using the reflexive property.

Therefore by AA similarity theorem , ΔACB ≈ ΔDCE

• AA similarity theorem says that if in two triangles the two pairs of corresponding angles are congruent then the triangles are similar .

Answer 2

Answer:

AA

Step-by-step explanation: