Question

$△ABC$ is an isosceles triangle in which $AB=AC$. Sides $BA$ is produced to $D$ such that $AD=AB$. Show that $\angle BCD$ is a right angle.

Answer 1

In $△ABC$, we have

$AB=AC$ $\mid$ given

$\angle ACB=\angle ABC$ ... (1) $\mid$ Since angles opp. to equal sides are equal

Now, $AB=AD$ $\mid$ Given

$\therefore AD=AC$ $\mid$ Since $AB=AC$

Thus , in $△ADC$, we have

$AD=AC$

$\Rightarrow \angle ACD=\angle ADC$ ... (2) $\mid$ Since angles opp. to equal sides are equal

Adding (1) and (2) , we get

$\angle ACB+\angle ACD=\angle ABC+\angle ADC$

$\Rightarrow \angle BCD=\angle ABC+\angle BDC$ $\mid$ Since$\angle ADC=\angle BDC$

$\Rightarrow \angle BCD+\angle BCD=\angle ABC+\angle BDC+\angle BCD$ $\mid$ Adding $\angle BCD$ on both sides

$\Rightarrow 2\angle BCD={180}^{\circ }$ $\mid$ Angle sum property

$\Rightarrow \angle BCD={90}^{\circ }$

Hence, $\angle BCD$ is a right angle.