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Let ABC be a triangle with incentre I. I...

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)`

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In `Delta APB, sin.(B)/(2) = (AP)/(AB)`

In `Delta AQC, sin.(C)/(2) = (AQ)/(AC)`
Now, in `Delta ABI`, using sine rule, we get
`(BI)/((sin.(A)/(2)) = (AB)/((cos.(C)/(2))`
And in `Delta ACI`, using sine rule, we get
`(CI)/((sin.(A)/(2)) = (AC)/(cos.(B)/(2))`
`:. (AP)/(BI) + (AQ)/(CI) = (AB sin.(B)/(2) cos.(C)/(2))/(AB sin.(A)/(2)) + (AC sin.(C)/(2) cos.(B)/(2))/(AC sin.(A)/(2))`
`= (sin((B + C)/(2)))/(sin.(A)/(2)) = cot.(A)/(2)`
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