Question

In a $\Delta ABC$, if point D is on the BC such that $\frac{AB}{AC}=\frac{BD}{DC}$ and $\angle B={70}^{\circ },\angle C={50}^{\circ }$.
Find $\angle BAD$.

Answer 1

We know that if a line through one of the vertex of a triangle divides the opposite side in the ratio of the other two sides, the line bisects the angle at the vertex.

So in the given equation,

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

$\Rightarrow AD$ bisects $\angle A$.

$\angle A={180}^{\circ }-({70}^{\circ }+{50}^{\circ })$ (Sum of angles of a triangle $={180}^{\circ }$)

$\therefore \angle BAD=\frac{\angle A}{2}=\frac{{60}^{\circ }}{2}={30}^{\circ }$