Question

In the given figure, AB =AC. D is a point on AC and E on AB such that AD =ED = EC = BC. Find $\frac{\angle B}{\angle A}$.

Answer 1

Let the base angle of $△ACD$ be $\angle A.$

$\angle D=180-2\angle A$

$\angle E=\angle A$

Consider $△DEC$

$\angle EDC=180-180-2\angle A$

$⟹\angle EDC=-2\angle A$

$\angle DEC=180-2\angle A-2\angle A=180-4\angle A$

$\angle CEB=180-180+4\angle A-\angle A$

$\therefore \angle B=\angle CEB=3\angle A$

$\therefore \frac{\angle B}{\angle A}=\frac{3}{1}$