Question

In the given figure, the ratio $\angle \mathrm{ABD}:\angle \mathrm{ACD}$ is

• A. $1:1$
• B. $2:1$
• C. $1:2$
• D. $2:3$

Answer 1

The correct option is A

$1:1$

Step-1: Find the relation between angles $\mathbf{(}\mathbf{\angle }\mathbf{B}$ and $\mathbf{\angle }\mathbf{C}\mathbf{)}$ in $\mathbf{△}\mathbf{ABC}$ and $\mathbf{△}\mathbf{BCD}$.

In $△\mathrm{ABC}$,

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

$\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{ACB}$ $($$\therefore$ Angles opposite to equal sides of the triangle are equal$)$

Let $\angle \mathrm{ABC}=\angle \mathrm{ACB}=\mathrm{p}$

In $△\mathrm{BCD}$,

$\mathrm{BD}=\mathrm{DC}$

$\Rightarrow \angle \mathrm{DBC}=\angle \mathrm{DCB}$ $($$\therefore$ Angles opposite to equal sides of the triangle are equal$)$

Let $\angle \mathrm{DBC}=\angle \mathrm{DCB}=\mathrm{q}$

Step-2: Find the ratio between $\mathbf{\angle }\mathbf{ABD}$ and $\mathbf{\angle }\mathbf{ACD}$

$\angle \mathrm{ABD}=\angle \mathrm{ABC}-\angle \mathrm{DBC}$

$\Rightarrow \angle \mathrm{ABD}=\mathrm{p}-\mathrm{q}......\left(1\right)$

$\angle \mathrm{ACD}=\angle \mathrm{ACB}-\angle \mathrm{DCB}$

$\Rightarrow \angle \mathrm{ACD}=\mathrm{p}-\mathrm{q}......\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$

$\Rightarrow \frac{\angle \mathrm{ABD}}{\angle \mathrm{ACD}}=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{p}-\mathrm{q}}$

$\Rightarrow \frac{\angle \mathrm{ABD}}{\angle \mathrm{ACD}}=\frac{1}{1}$

Hence, $\angle \mathrm{ABD}:\angle \mathrm{ACD}=1:1$

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