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 = \angle a \\) \\( \angle a = \angle d \\) \\( \angle a = \angle acb \\) \\( \angle a = \angle abc \\)

Explanation:

To prove similarity by SAS (Side - Angle - Side), we need two sides in proportion and the included angle equal. Here, the sides \(\frac{AE}{AC}=\frac{AD}{AB}\) are in proportion. The included angle for both triangles \(\triangle AED\) and \(\triangle ACB\) is \(\angle A\). By the reflexive property of angles, \(\angle A=\angle A\).

Answer:

\(\angle A=\angle A\)