Question

The incircle touches side $BC$ of triangle $ABC$ at $D$ and $ID$ is produced to $H$ so that $DH=s,$ where $s$ and $I$ are the semi-perimeter and incentre of triangle $ABC$ respectively. If $HBIC$ is cyclic, then $\cot \left(\frac{A}{4}\right)$ is equal to

• A. $2+\sqrt{3}$
• B. $2-\sqrt{3}$
• C. $\sqrt{2}+1$
• D. $\sqrt{2}-1$

Answer 1

The correct option is C $\sqrt{2}+1$


As $HBIC$ is cyclic, so
$r\cdot s=(s-b)(s-c)$
$\Rightarrow △=(s-b)(s-c)$
$\Rightarrow \frac{1}{2}bc\sin A=(s-b)(s-c)$
$\Rightarrow \frac{1}{2}\sin A=\frac{(s-b)(s-c)}{bc}={\sin }^2\frac{A}{2}$
$\Rightarrow \sin A=2{\sin }^2\frac{A}{2}$
$\Rightarrow 2\sin \frac{A}{2}\cos \frac{A}{2}=2{\sin }^2\frac{A}{2}$
$\Rightarrow \sin \frac{A}{2}(\cos \frac{A}{2}-\sin \frac{A}{2})=0$
$\Rightarrow \cot \frac{A}{2}=1,$ so $\frac{A}{2}=\frac{\pi }{4}$
$\therefore \frac{A}{4}=\frac{\pi }{8}$
Hence, $\cot \frac{A}{4}=\cot \frac{\pi }{8}=\sqrt{2}+1$