Question:
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°
Solution:
(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 $\triangle A B C$,
$\angle A+\angle B+\angle C=180^{\circ}$
$\Rightarrow \angle A=180-\angle B-\angle C$
$\Rightarrow \angle A=180-70-50=60^{\circ}$
$\because \angle B A D=\angle C A D=\frac{1}{2} \angle B A C$
$\therefore \angle B A D=\frac{1}{2} \times 60=30^{\circ}$