Question

$△ABC$ is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. $\angle BCD$ is equal to

• A. ${90}^{\circ }$
• B. ${45}^{\circ }$
• C. ${37}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

Given, AB = AC and BA = AD.

In $△ABC$,

$\angle ABC=\angle ACB=x$ (angles opposite to equal sides)

In $△ACD$,

$\angle ADC=\angle ACD=y$ (angles opposite to equal sides)

$\therefore \angle BCD=x+y$

In $△ABC$,

$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$ (angle sum property of a triangle)

$x+x+\angle BAC={180}^{\circ }$

$\angle BAC={180}^{\circ }-2x$

$\angle BAC+\angle DAC={180}^{\circ }$ (Linear pair)

$({180}^{\circ }-2x)+\angle DAC={180}^{\circ }$

So, $\angle DAC=2x$

In $△DAC$,

$\angle DAC+\angle ADC+\angle ACD={180}^{\circ }$ (angle sum property of a triangle)

$2x+y+y={180}^{\circ }$

$2x+2y={180}^{\circ }$

$x+y={90}^{\circ }$

$\angle BCD=\angle BCA+\angle ACD$
$=x+y$
$={90}^{\circ }$

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