Read the proof. Given: AB ∥ DE Prove: △ACB ~ △DCE We are given AB ∥ DE. Because the lines are parallel and segment CB crosses bo
th 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.
2 answers:
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.
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