Question

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.

Answer 1

Triangles ACB and DCE are considered similar due to the Angle-Angle (AA) Similarity Theorem, as two pairs of angles are congruent, given that AB is parallel to DE and CB is a transversal.

Step-by-step explanation:

The student is discussing the conditions required to prove that two triangles are similar. If AB ∥ DE and segment CB is a transversal, ∠CED ≅ ∠CBA are corresponding angles, and are therefore congruent. Additionally, by the reflexive property, we can state ∠C ≅ ∠C. The combination of these angle congruences suggests that the triangles △ACB and △DCE are similar by the AA (Angle-Angle) Similarity Theorem, because two pairs of angles in the triangles are congruent. The SSS (Side-Side-Side) similarity theorem wouldn't apply here because we do not have information about the sides' proportions, nor would the AAS (Angle-Angle-Side) or ASA (Angle-Side-Angle) similarity theorems, since we haven't established a congruence of a side along with the angles.

Answer 2

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.