Question

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

Answer 1

Since $AB=AC$ and $AD=AB$, therefore $AC=AD$. Thus
$\angle ABC=\angle ACB$ .....i

$\angle ADC=\angle ACD$ .....ii

Now, in $\Delta BDC$
$\angle BDC+\angle DBC+\angle BCD={180}^{\circ }$ [ Angle sum property]

$\Rightarrow \angle ACD+\angle ACB+\angle ACD+\angle ACB={180}^{\circ }$ [Using (i) and (ii)]

$\Rightarrow 2\angle ACD+2\angle ACB={180}^{\circ }$

$\Rightarrow \angle ACD+\angle ACB=\frac{{180}^{\circ }}{2}$

$\Rightarrow \angle BCD={90}^{\circ }$ [$\angle BCD=\angle ACD+\angle ACB$]

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