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.

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.

Mathematics · High School · Mon Jan 18 2021

llm

Answered on Sat Nov 18 2023

Given: AB ∥ DE Prove: △ACB ~ △DCE

The given statement says that line segment AB is parallel to line segment DE. Since we are trying to prove that triangles ACB and DCE are similar, we have to show that they have the same shape but not necessarily the same size. To prove similarity, we can use several theorems, but the most appropriate one based on the given information would be the AA (Angle-Angle) similarity theorem.

We are given that AB ∥ DE and segment CB intersects both AB and DE. This means that CB acts as a transversal for the parallel lines AB and DE.

By the corresponding angles postulate, if a transversal intersects two parallel lines, the pairs of corresponding angles are congruent. Therefore, ∠CED and ∠CBA are corresponding angles and since AB is parallel to DE, these angles are congruent (∠CED ≅ ∠CBA).

Additionally, we can state ∠C ≅ ∠C for both triangles using the reflexive property (which simply states that any geometric figure is congruent to itself). Both triangles share angle C.

With two pairs of angles proven to be congruent (∠CBA ≅ ∠CED and ∠C ≅ ∠C), the AA similarity theorem states that two triangles are similar if two pairs of corresponding angles are congruent.

Thus, by the AA similarity theorem, we can conclude that △ACB ~ △DCE.

In this case, the other theorems (SSS, AAS, ASA) mentioned are not appropriate because we haven’t established the necessary side lengths relationships for SSS (Side-Side-Side) or the angle-side-angle relationships for AAS (Angle-Angle-Side) and ASA (Angle-Side-Angle). We strictly used angles to establish the similarity of the triangles.

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