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

A.AA similarity theorem.B.SSS similarity theorem.C.AAS similarity theorem.D.ASA similarity theorem.

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06-06-2017
Mathematics

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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

A.AA similarity theorem.B.SSS similarity theorem.C.AAS similarity theorem.D.ASA similarity theorem.

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nobillionaireNobley nobillionaireNobley · Answer 1 · 2017-06-19T06:02:12+02:00

Given that 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.

From the proof, we have been able to establish two congruencies, that is that angles CED and CBA are congruent and angle C is congruent to angle C.

Therefore, we conclude that △ACB ~ △DCE by the AA similarlity theorem.

syanne12 syanne12 · Answer 2 · 2017-06-19T15:10:31+02:00

syanne12 syanne12

19-06-2017

I believe that the answer will be A.

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