Question

In the given figure, $△$ ABC is traingle such that $\frac{AB}{AC}=\frac{BD}{DC}$, $\angle$ B = 70, $\angle$ C = 50. Find $\angle$ BAD.

Answer 1

If a line through one vertex of a triangle divides the opposite side in the ratio of the other two sides, then the line bisects the angle at the vertex.

Therefore $\angle 1=\angle 2$

So in $△ABC$ by triangle sum property:

$\angle 1+\angle 2+\angle ABC+\angle ACB={180}^{\circ }$

$2\angle 1+\angle ABC+\angle ACB={180}^{\circ }$

Given: $\angle ABC={70}^{\circ },\angle ACB={50}^{\circ }$

$2\angle 1+{70}^{\circ }+{50}^{\circ }={180}^{\circ }$

$2\angle 1={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$\angle 1=\angle BAD={30}^{\circ }$

So $\angle BAD={30}^{\circ }$