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Let A B C be a triangle with incentre at...

Let `A B C` be a triangle with incentre at `Idot` Also, let `Pa n dQ` be the feet of perpendiculars from `AtoB Ia n dC I ,` respectively. Then which of the following results are correct? `(A P)/(B I)=(sinB/2cosC/2)/(sinA/2)` (b) `(A Q)/(C I)=(sinC/2cosB/2)/(sinA/2)` `(A P)/(B I)=(sinC/2cosB/2)/(sinA/2)` (d) `(A P)/(B I)+(a Q)/(C I)=sqrt(3)if/_A=60^0`

Text Solution

Verified by Experts

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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