Question

Question 6
$\Delta ABC$ is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that $\angle BCD$ is a right angle.

Answer 1

$In-\Delta ABCAB=AC\left(given\right)\Rightarrow \angle ACB=\angle ABC\left(anglesoppositetoequalsidesofatrianglearealsoequal\right)NowIn-\Delta ACDAC=AD\angle ADC=\angle ACD\left(anglesoppositetoequalsidesofatrianglearealsoequal\right)Now,in\Delta BCD\angle ABC+\angle BCD+\angle ADC={180}^{\circ }\left(anglesumpropertyofatriangle\right)\Rightarrow \angle ABC+\angle ACB+\angle ACD+\angle ACD={180}^{\circ }(\because \angle BCD=\angle ACB+\angle ACD)\Rightarrow \angle ACB+\angle ACB+\angle ACD+\angle ACD={180}^{\circ }(\because \angle ABC=\angle ACB)\Rightarrow 2(\angle ACB+\angle ACD)={180}^{\circ }\Rightarrow 2\left(\angle BCD\right)={180}^{\circ }\Rightarrow \angle BCD={90}^{\circ }$