Question

In triangle $ABC$ if $\angle A={60}^0$ and $AD$ is the angle bisector of $\angle A$, $D$ is point on $BC$, then length $AD$ is equal to

• A. $\frac{bc}{b+c}$
• B. $\frac{2bc}{b+c}$
• C. $\frac{\sqrt{3}bc}{b+c}$
• D. None of these

Answer 1

The correct option is C $\frac{\sqrt{3}bc}{b+c}$

Area of $△ABC=$ Area of $△ABD+$ Area of $△ADC$

$\Rightarrow \frac{1}{2}bc\sin {60}^0=\frac{1}{2}c.AD\sin {30}^0+\frac{1}{2}b.AD\sin {30}^0$

$\Rightarrow bc\frac{\sqrt{3}}{2}=\frac{1}{2}(b+c)AD\Rightarrow AD=\frac{\sqrt{3}bc}{b+c}$