Question

In the figure, $AD:DC=2:3$, then $\angle ABC$ is equal to:

• A. ${30}^o$
• B. ${40}^o$
• C. ${45}^o$
• D. ${110}^o$

Answer 1

The correct option is B ${40}^o$

Given: $\frac{AD}{DC}=2:3$
$AB=10$ and $BC=15$
$\frac{AB}{BC}=\frac{10}{15}=\frac{2}{3}$
hence, $\frac{AB}{BC}=\frac{AD}{DC}=\frac{2}{3}$
Thus, by converse of Angle Bisector theorem,
BD bisects $\angle ABC$

Now, In $△BDC$
Sum of angles = 180
$\angle BDC+\angle DBC+\angle BCD=180$
$130+30+\angle DBC=180$
$\angle DBC={20}^{\circ }$
Since, DB bisecs $\angle B$
Hence, $\angle B=2\angle DBC=2\times 20={40}^{\circ }$