Question

$I$ is the incenter of $△ABC$. IF $P$ and $Q$ be the feet of perpendicular from $A$ to $BI$ and $CI$, respectively, then $\frac{AP}{BI}+\frac{AQ}{CI}=$

• A. $\cot \frac{A}{2}$
• B. $\tan \frac{A}{2}$
• C. $\sin \frac{A}{2}$
• D. $\cos \frac{A}{2}$

Answer 1

The correct option is A $\cot \frac{A}{2}$

In $△APB,\sin \frac{B}{2}=\frac{AP}{AB}$

n $△AQC,\sin \frac{C}{2}=\frac{AQ}{AC}$

Now, In $△ABI$, we have

$\frac{BI}{\sin \frac{A}{2}}=\frac{AB}{\sin (\pi -\frac{A}{2}-\frac{B}{2})}=\frac{AB}{\cos \frac{C}{2}}$

and in $△ACI$, we have $\frac{CI}{\sin \frac{A}{2}}=\frac{AC}{\cos \frac{B}{2}}$

So, $\frac{AP}{BI}+\frac{AQ}{CI}=\frac{\sin \frac{B}{2}\cos \frac{C}{2}}{\sin \frac{A}{2}}+\frac{\sin \frac{C}{2}\cos \frac{B}{2}}{\sin \frac{A}{2}}$

$=\frac{\sin (\frac{B}{2}+\frac{C}{2})}{\sin \frac{A}{2}}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\cot \frac{A}{2}$