In ∆ABC, it is given that


In $\triangle A B C$, it is given that $\frac{A B}{A C}=\frac{B D}{D C}$. If $\angle B=70^{\circ}$ and $\angle C=50^{\circ}$, then $\angle B A D=$ ?

(a) 30°
(b) 40°
(c) 45°
(d) 50°



(a) 30°">°°
We have:

$\frac{A B}{A C}=\frac{B D}{D C}$

Applying angle bisector theorem, we can conclude that $A D$ bisects $\angle A$.

In $\triangle A B C$,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \angle A=180-\angle B-\angle C$

$\Rightarrow \angle A=180-70-50=60^{\circ}$

$\because \angle B A D=\angle C A D=\frac{1}{2} \angle B A C$

$\therefore \angle B A D=\frac{1}{2} \times 60=30^{\circ}$


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