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For triangle `ABC,R=5/2 and r=1.` Let `I` be the incenter of the triangle and `D,E and F` be the feet of the perpendiculars from `I->BC,CA and AB,` respectively. The value of `(IDxIExIF)/(IAxIBxIC)` is equal to (a) `5/2` (b) `5/4` (c) `10` (d) `1/5`

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Best answer
In`/_AIF` and `/_AIE`
`(IF)/(sin(A/2))=AI=(IE)/(sin(A/2))`
`(AI)^2=(IE*IF)/(sin^2(A/2))`
`(BI)^2=(IF*ID)/(sin^2(B/2))`
`(CCI)^2=(IE*ID)/(sin(C/2))`
`(AI)^2*(BI)^2*(CI)^2=((IE)^2(IF)^2(ID)^2)/(sin^2(A/2)sin^2(B/2)sin^2(C/2))`
`(IE*IF*ID)/(AI*BI*CI)=sin(A/2)sin(B/2)sin(C/2)=r/(4R)`
`r=4Rsin(A/2)sin(B/2)sin(C/2)`
`r=1/10`
Option C is correct.

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