Question

refer to the diagram shown. if ∠bac≅∠dca, the choose theorem can be used to show that △abe≅△cde

Explanation:

Step1: Identify vertical angles

$\angle AEB=\angle DEC$ (vertical - angles are congruent).

Step2: Given angle equality

We are given that $\angle BAC=\angle DCA$.

Step3: Identify side - angle - side (SAS), angle - side - angle (ASA), or angle - angle - side (AAS) criteria

Since we have two pairs of angles equal and the included sides are also equal (because they are vertical - angle pairs), the Angle - Angle - Side (AAS) theorem can be used. In $\triangle ABE$ and $\triangle CDE$, we have $\angle BAC=\angle DCA$, $\angle AEB = \angle DEC$, and $AE = EC$ (not given explicitly but can be inferred from the vertical - angle and congruence setup in a parallelogram - like figure where the diagonals bisect each other in some cases, and the non - included sides are considered).

Answer:

AAS