Question

In an isosceles triangle $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{BA}$ is produced to $\mathrm{D}$, such that $\mathrm{AB}=\mathrm{AD}$ then $\angle \mathrm{BCD}$ is

• A. $70°$
• B. $90°$
• C. $60°$
• D. $45°$

Answer 1

The correct option is B

$90°$

Step-1: Draw an isosceles triangle $\mathbf{AB}\mathbf{}\mathbf{=}\mathbf{}\mathbf{AC}$ and $\mathbf{BA}$ is produced to $\mathbf{D}$, such that $\mathbf{AB}\mathbf{}\mathbf{=}\mathbf{}\mathbf{AD}$.

In $∆\mathrm{ABC}$,

$\Rightarrow \mathrm{AB}=\mathrm{AC}$,

So, $\angle \mathrm{ABC}=\angle \mathrm{ACB}.....\left(1\right)$ (Angles opposite to equal sides are equal)

$\Rightarrow \mathrm{AB}=\mathrm{AD}$ then, $\mathrm{AC}=\mathrm{AD}$
So, $\angle \mathrm{ACD}=\angle \mathrm{ADC}.....\left(2\right)$ (Angles opposite to equal sides are equal)

Step-2: Find $\mathbf{\angle }\mathbf{BCD}$

Adding $\left(1\right)$and $\left(2\right)$

$\Rightarrow$$\angle \mathrm{ABC}+\angle \mathrm{ADC}=\angle \mathrm{ACB}+\angle \mathrm{ACD}$

$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{ADC}=\angle \mathrm{BCD}.....\left(3\right)$

In $∆\mathrm{BCD}$
$\angle \mathrm{ABC}+\angle \mathrm{BCD}+\angle \mathrm{BDC}=180°$

$\Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BDC}=180°-\angle \mathrm{BCD}.....\left(4\right)$

From $\left(4\right)$ and $\left(3\right)$

$\Rightarrow 180°-\angle \mathrm{BCD}=\angle \mathrm{BCD}$

$\Rightarrow 2\angle \mathrm{BCD}=180°$

$\Rightarrow \angle \mathrm{BCD}=90°$
Hence, $\angle \mathrm{BCD}=90°$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.