Question

Let $\Delta ABC$ be a triangle with incentre at $I$. Also let $P$ and $Q$ be the feet of perpendiculars from $A$ to $BI$ and $CI$, respectively. Then which of the following results are correct ?

• A. $\frac{AQ}{CI}=\frac{\sin \left(\frac{C}{2}\right)\cos \left(\frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
• B. $\frac{AP}{BI}=\frac{\sin \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
• C. $\frac{AQ}{CI}=\frac{\cos \left(\frac{C}{2}\right)\sin \left(\frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
• D. $\frac{AP}{BI}+\frac{AQ}{CI}=\sqrt{3}$ if $\angle A={30}^{\circ }$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{AQ}{CI}=\frac{\sin \left(\frac{C}{2}\right)\cos \left(\frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
B $\frac{AP}{BI}=\frac{\sin \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

From $\Delta APB$,
$AP=AB\sin \left(\frac{B}{2}\right)$
From $\Delta AIB$,

$\frac{BI}{AB}=\frac{\sin \left(\frac{A}{2}\right)}{\cos \left(\frac{C}{2}\right)}$
$(A+B+C=\pi )\to (\frac{A+B+C}{2}=\frac{\pi }{2})\Rightarrow (\angle AIB=\frac{\pi +C}{2})$
$\Rightarrow \frac{AP}{BI}=\frac{\sin \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Similarly, $\frac{AQ}{CI}=\frac{\sin \left(\frac{C}{2}\right)\cos \left(\frac{B}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\Rightarrow \frac{AP}{BI}+\frac{AQ}{CI}=\frac{\sin \left(\frac{B+C}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\cot \left(\frac{A}{2}\right)$
$\because \cot \left(\frac{A}{2}\right)=\sqrt{3}\Rightarrow \angle A={60}^{\circ }$