Question

to prove that $\triangle aed \sim \triangle acb$ by sas, jose shows that $\frac{ae}{ac} = \frac{ad}{ab}$. jose also has to state that $\angle a \cong \angle abc$. $\angle a \cong \angle acb$. $\angle a \cong \angle a$. $\angle a \cong \angle d$.

Explanation:

Step1: Recall SAS Similarity Criterion

The SAS (Side - Angle - Side) similarity criterion for triangles states that if two sides of one triangle are proportional to two sides of another triangle, and the included angles are congruent, then the two triangles are similar.

Step2: Identify the Included Angle

For $\triangle AED$ and $\triangle ACB$, we have the proportion $\frac{AE}{AC}=\frac{AD}{AB}$. The included angle between sides $AE$ and $AD$ in $\triangle AED$ and between sides $AC$ and $AB$ in $\triangle ACB$ is $\angle A$. So, we need to state that $\angle A\cong\angle A$ (by the reflexive property of congruence) to satisfy the SAS similarity criterion.

Answer:

$\angle A\cong\angle A$ (the option with $\angle A\cong\angle A$)