Question

$△ABC$ is an isosceles $△$ with $AB=AC$ side $BA$ is produced to $D$ such that $AB=AD$ prove $\angle BCD$ is a right angle.

Answer 1

$AB=AC$

$\Rightarrow \angle B=\angle ACB=\theta$ (say)

Similarly $AB=AD=AC$

$\Rightarrow \angle ACD=\angle ADC=\theta$

$\angle BAC={180}^{\circ }-2\theta$(Consider the $△$ ABC)

$\angle CAD={180}^{\circ }-2\theta$(Consider the $△$ ADC)

Since $BAD$ is a straight line

$\angle BAC+\angle CAD={180}^{\circ }\Rightarrow {360}^{\circ }-4\theta ={180}^{\circ }\Rightarrow 4\theta ={180}^{\circ }\Rightarrow \theta ={45}^{\circ }\therefore \angle BCD=2\theta ={90}^{\circ }$