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.
2 answers:
Answer: AAA similarity.
Step-by-step explanation: CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have
∠CED ≅ ∠CBA,
∠CDE ≅ ∠CAB
and
∠DCE ≅ ∠ACB [same angle]
Hence, by AAA (angle-angle-angle) similarity,
△CED ~ △ABC.
Thus, the correct option is AAA similarity.
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