Question

In an isosceles triangle AB = AC and BA is produced to D, such that AB = AD. What is the value of $\angle BCD$?

• A. ${70}^{\circ }$
• B. ${90}^{\circ }$
• C. ${60}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B

${90}^{\circ }$

Given, AB = AC and AD = AB($⟹$ AD = AC)
Since angles opposite to equal sides are equal, we must have

$\angle ABC=\angle ACB\dots \left(i\right)$ and
$\angle ADC=\angle ACD\dots \left(ii\right)$


Adding (i) and (ii), we get
$\angle ABC+\angle ADC=\angle ACB+\angle ACD$

$\angle BCD=\angle ACB+\angle ACD$
$\Rightarrow \angle BCD=\angle ABC+\angle ADC$ ————— (iii)
In $\Delta BCD$,
$\angle BCD+\angle DBC+\angle BDC={180}^{\circ }$ ————— (iv) [Angle sum property]

$\angle DBC$ is same as $\angle ABC$ and $\angle ACD=\angle ADC$
From (iii) and (iv), we get
$2\angle BCD={180}^{\circ }$

$\Rightarrow \angle BCD=\frac{{180}^{\circ }}{2}={90}^{\circ }$