Question

In the adjoining figure, AB = 10 cm, BC =15 cm AD : DC = 2 : 3, then $\angle$ABC is equal to -

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

Answer 1

The correct option is B

${40}^{\circ }$

Clearly, $\frac{AD}{DC}$ = $\frac{2}{3}$ and $\frac{AB}{BC}$ = $\frac{10}{15}$ = $\frac{2}{3}$

So, $\frac{AD}{DC}$ = $\frac{AB}{BC}$

Thus, BD divides AC in the ratio of the other two sides.

By Converse of Angular Bisector theorem,
BD is the bisector of $\angle$B
$\therefore$ $\angle$ABC = 2 ( $\angle$CBD)

Now, $\angle$CBD = ${180}^{\circ }$ - (${130}^{\circ }$ + ${30}^{\circ }$) .... (Angle Sum Property)
= ${20}^{\circ }$

But , $\angle$ABC = 2 ( $\angle$CBD)
= 2 x 20
= ${40}^{\circ }$