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

A

Explanation:

Answer 2

(A) AA Similarity Theorem

Explanation:

Given: AB ∥ DE

To Prove:
`\triangle ACB \sim \triangle DCE`

Given Triangle ABC with Line DE drawn inside of the triangle and parallel to side AB. The line DE forms a new triangle DCE.

Because AB∥DE 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,
`\triangle ACB \sim \triangle DCE` by the AA similarity theorem.

Remark: In the diagram, we can see that the two triangles share Angle C and have two equal angles at E and B. Therefore, they are similar by the Angle-Angle Similarity Theorem.