Question

to prove that δaed ~ δacb by sas, jose shows that (\frac{ae}{ac} = \frac{ad}{ab}). jose also has to state that ∠a=∠a ∠a=∠d ∠a=∠acb ∠a=∠abc

Explanation:

To prove two triangles similar by SAS (Side - Angle - Side) similarity criterion, we need two pairs of corresponding sides in proportion and the included angle equal.

For $\triangle AED$ and $\triangle ACB$, we have $\frac{AE}{AC}=\frac{AD}{AB}$. The included angle between the sides $AE,AD$ in $\triangle AED$ and the sides $AC,AB$ in $\triangle ACB$ is $\angle A$. So, the angle that needs to be equal is $\angle A=\angle A$ (common angle).

• Option $\angle A = \angle D$: $\angle D$ is not the included angle for the proportional sides in $\triangle AED$ and $\triangle ACB$.
• Option $\angle A=\angle ACB$: $\angle ACB$ is an angle in $\triangle ACB$ but not the included angle for the proportional sides.
• Option $\angle A=\angle ABC$: $\angle ABC$ is an angle in $\triangle ACB$ but not the included angle for the proportional sides.

Answer:

$\boldsymbol{\angle A=\angle A}$ (the first option: $\boldsymbol{\angle A=\angle A}$)