MCQ
In $\triangle ABC$ it is given that $\frac{ AB }{ AC }=\frac{ BD }{ DC }$. If $\angle B =70^{\circ}$ and $\angle C =50^{\circ}$ then $\angle BAD =$ ?
- ✓$30^{\circ}$
- B$40^{\circ}$
- C$45^{\circ}$
- D$50^{\circ}$
✓
Answer
Correct option: A.
$30^{\circ}$
$\angle A=180^{\circ}-\left(70^{\circ}+50^{\circ}\right)=180^{\circ}-120^{\circ}=60^{\circ}$
Since $\frac{B D}{D C}=\frac{A B}{A C}$, it means $A D$ is the bisector of $\angle A$.
$\therefore \angle BAD=\frac{1}{2} \times \angle A$
$=\frac{1}{2} \times 60^{\circ}$
$=30^{\circ}$
Since $\frac{B D}{D C}=\frac{A B}{A C}$, it means $A D$ is the bisector of $\angle A$.
$\therefore \angle BAD=\frac{1}{2} \times \angle A$
$=\frac{1}{2} \times 60^{\circ}$
$=30^{\circ}$
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