Question

$\Delta ABC$ is an isosceles triangle in which$\mathit{}AB\mathit{=}AC$. Side $BA$is produced to $D$ such that $AD=AB$ (see Fig.) . Show that $\angle BCD$is a right angle.

Answer 1

Showing that $\angle BCD$is a right angle:

Given:

$AB=AC$ and$AD=AB$

In $∆ABC,$

$\begin{array}{rcl}AB& =& AC\\ \angle ACB& =& \angle ABC——————–\left(i\right)[Anglesoppositetoequalsidesareequal]\end{array}$

In$∆ACD,$

$\begin{array}{rcl}AC& =& AD\\ \angle ACD& =& \angle ADC——————–\left(ii\right)[Anglesoppositetoequalsidesareequal]\end{array}$

On adding equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ we get

$\begin{array}{rcl}\angle ACB+\angle ACD& =& \angle ABC+\angle ADC\\ \angle BCD& =& \angle ABC+\angle BDC\end{array}$

Adding $\angle \mathrm{BCD}$ on both side

$\begin{array}{rcl}\angle BCD+\angle BCD& =& \angle ABC+\angle BDC+\angle BCD\end{array}$

$\begin{array}{rcl}2\angle BCD& =& 180°(Byanglesumproperty)\\ & \Rightarrow & \angle BCD=\frac{180°}{2}\\ & \Rightarrow & \angle BCD=90°\end{array}$

Hence we proved that $\angle BCD$is a right angle: