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.

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.

poojatomarb76 poojatomarb76 · Answer 2 · 2022-07-04T11:03:52+02:00

AAA similarity is the right answer.AAA similarity shows the similarity of angles.

What are the conditions of the congruent triangle?

Triangles that are equivalent in terms of size and form are known as a congruent triangle.

Given;

AB ∥ DE

Since CB is the transversal for the lines AB and DE, we have CED and CBA by virtue of the transverse property.

Similar to how AB and DE work as transversals for one another, CA does the same, and by applying the same attribute, we may get CDE CAB. Currently, we have triangles CED and ABC.

∠CED ≅ ∠CBA,

∠CDE ≅ ∠CAB

Due to the similar angle;

∠DCE ≅ ∠ACB

We can obtain AAA (angle-angle-angle) similarity applied here.

Hence, by

△CED ~ △ABC.

Therefore, AAA similarity is the right answer.

To learn more about the congruent triangle refer;

https://brainly.com/question/12413243

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