Question

$△ABC$ is an isosceles triangle and AB = AC. If side BA is produced to D such that BA = AD, what is the measure of $\angle DCB$?

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is C

${90}^{\circ }$

$\text{In}△ABC,AB=AC.$
We know, angles opposite to equal sides of a triangle are equal.
$\text{So,}\angle ABC=\angle ACB=\angle 1\text{(say)}$
$\text{Given that, side BA is produced to D such that BA = AD.}$
$\text{But we already have AB=AC.}$
$\text{Then, in}△ACD,AD=AC.\text{So,}\angle ADC=\angle ACD=\angle 2\text{(say)}$

$\text{In}△ABC,\angle 1+\angle 1+\angle 3={180}^{\circ }\cdots \left(i\right)$
$\text{In}△ADC,\angle 2+\angle 2+\angle 4={180}^{\circ }\cdots \left(ii\right)$
$\text{Also},\angle 3+\angle 4={180}^{\circ }$ (Linear pair)
$\text{Adding}\left(i\right)\text{and}\left(ii\right),\text{we get}2(\angle 1+\angle 2)+(\angle 3+\angle 4)={360}^{\circ }2(\angle 1+\angle 2)={180}^{\circ }\angle 1+\angle 2={90}^{\circ }⟹\angle DCB={90}^{\circ }$