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Incircle of DeltaABC touches the sides B...

Incircle of `DeltaABC` touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question
The length of side EF is

A

`r sin.(A)/(2)`

B

`2r sin.(A)/(2)`

C

`r cos.(A)/(2)`

D

`2r cos.(A)/(2)`

Text Solution

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The correct Answer is:
D

Points I, D, C, E are concylic. Therefore,
`angleEID = pi -C`
Also, I is circumcenter of `DeltaDEF`. Hence,
`angleDFE = (1)/(2) angleDIE = (pi -C)/(2)`
Similarly, `angleFDE = (pi -A)/(2)`
`angleFED = (pi -B)/(2)`
`:.` Area of `DeltaDEF = 2r^(2) sin (angleD) sin (angleE) sin (angleF)`
`= 2r^(2) "cos"(A)/(2) "cos"(B)/(2) "cos"(C)/(2)`
Also by sine rule, in `DeltaDEF`,
`(EF)/(sin angleD) = 2r`
or `EF = 2r cos.(A)/(2)`
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