Question

Let $ABC$ be a triangle with $\angle BAC={120}^{\circ }$ and $\left(AB\right)\cdot \left(AC\right)=1.$ Also, let $AD$ be the length of the angle bisector of angle $A$ of the triangle. Then

• A. Minimum value of $AD$ is $\frac{1}{2}$
• B. Maximum value of $AD$ is $\frac{1}{2}$
• C. $AD$ is minimum when $△ABC$ is isosceles.
• D. $AD$ is maximum when $△ABC$ is isosceles.

Answer 1

Answer: (B), (D)

The correct options are
B Maximum value of $AD$ is $\frac{1}{2}$
D $AD$ is maximum when $△ABC$ is isosceles.


Let $AB=c=x$ and $AC=b=\frac{1}{x}$
$AD=\frac{2bc}{b+c}\cos \left(\frac{A}{2}\right)\Rightarrow AD=\frac{1}{x+\frac{1}{x}}$

$x+\frac{1}{x}\ge 2\Rightarrow \frac{1}{x+\frac{1}{x}}\le \frac{1}{2}\Rightarrow AD\le \frac{1}{2}$

Equality holds when $x=\frac{1}{x}=1$