Question

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

Answer 1

Given: $△ABC$ is isosceles and $AB=AC$.

Also, $AD=AB$

In $\Delta ABC$

$AB=AC$

By isosceles triangle theorem,

$\angle ABC=\angle ACB$ .....(1)

In $\Delta ADC$

By isosceles triangle theorem,

$AB=AD=AC$ ......(given)

$\therefore \angle ADC=\angle ACD$ ....(2)

In $\Delta BCD$,

$\angle ABC+\angle BCD+\angle BDC={180}^0$ ...Angle sum property

$\Rightarrow (\angle ACB+\angle ACD)+\angle BCD={180}^0$ ..From (1) and (2)

$\Rightarrow \angle BCD+\angle BCD={180}^0$

$\Rightarrow 2\angle BCD={180}^0$

$\Rightarrow \angle BCD={90}^0$ $\left[henceproved\right]$