Let ABC be a triangle with incentre I. If P and Q are the feet of the perpendiculars from A to BI and CI, respectively, then prove that `(AP)/(BI) + (AQ)/(Cl) = cot.(A)/(2)`

Options

A. `(AP)/(Bl)=(sinB//2cosC//2)/(sinA//2)`
B. `(AQ)/(Cl)=(sinC//2cosB//2)/(sinA//2)`
C. `(AP)/(Bl)=(sinC//2cosB//2)/(sinA//2)`
D. `(AP)/(Bl)+(AQ)/(Cl)=sqrt(3)` if `/_A=60^(@)`

Answer

A, B, D

Explanation

In `/_\APB`, we have `AP=ABsinB//2`
In `/_\AIB`, we have `(BI)/(AB)=(sin(A//2))/(cos(C//2))`
`implies (AP)/(BI)=(sinB//2cosC//2)/(sinA//2)`
Similarly, `(AQ)/(Cl)=(sinC//2cosB//2)/(sinA//2)`
Now, `(AP)/(BI)+(AQ)/(Cl)=(sin((B+C)/2)=cot(A//2)`
`cot(A//2)=sqrt(3), /_A=60^(@)`