Question

to prove that \\( \triangle aed \sim \triangle acb \\) by sas, jose shows that \\( \frac { a e } { a c } = \frac { a d } { a b } \\) jose also has to state that \\( \angle a = \angle a \\) \\( \angle a = \angle d \\) \\( \angle a = \angle a c b \\) \\( \angle a = \angle a b c \\)

Explanation:

To prove two triangles similar by SAS (Side - Angle - Side) similarity criterion, we need two pairs of sides in proportion and the included angle equal. Here, we already have \(\frac{AE}{AC}=\frac{AD}{AB}\). The included angle for both ratios in \(\triangle AED\) and \(\triangle ACB\) is \(\angle A\). So, we need to state that \(\angle A=\angle A\) (common angle).

Answer:

\(\angle A = \angle A\)