Question

In ∆ABC, it is given that $\frac{AB}{AC}=\frac{BD}{DC}.\mathrm{If}\angle B=70°\mathrm{and}\angle C=50°,\mathrm{then}\angle BAD=?$


(a) 30°
(b) 40°
(c) 45°
(d) 50°

Answer 1

(a) 30$°$
We have:

$\frac{AB}{AC}=\frac{BD}{DC}$

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

$$\begin{gathered} \mathrm{In}△ABC, \\ \angle A+\angle B+\angle C=180° \\ \Rightarrow \angle A=180-\angle B-\angle C \\ \Rightarrow \angle A=180-70-50=60° \\ \because \angle BAD=\angle CAD=\frac{1}{2}\angle BAC \\ \therefore \angle BAD=\frac{1}{2}\times 60=30° \end{gathered}$$