In Fig. 4.58, ∆ABC is a triangle such that ABAC=BDDC,
Question:

In Fig. 4.58, $\triangle \mathrm{ABC}$ is a triangle such that $\mathrm{ABAC}=\mathrm{BDDC}, \angle \mathrm{B}=70^{\circ}, \angle \mathrm{C}=50^{\circ}$. Find the $\angle \mathrm{BAD}$.

Solution:

It is given that in $\triangle A B C, \frac{A B}{A C}=\frac{B D}{D C}, \angle B=70^{\circ}$ and $\angle C=50^{\circ}$.

We have to find $\angle B A D$.

In $\triangle A B C$,

$\angle A=180^{\circ}-\left(70^{\circ}+50^{\circ}\right)$

$=180^{\circ}-120^{\circ}$

$=60^{\circ}$

Since $\frac{A B}{A C}=\frac{B D}{D C}$, therefore, $A D$ is the bisector of $\angle \mathrm{A}$.

Hence, $\angle B A D=\frac{60^{\circ}}{2}=30^{\circ}$