Incircle of `DeltaABC` touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question
`angleDEF` is equal to

Options

A. `2r^(2) sin(2A) sin(2B) sin(2C)`
B. `2r^(2) cos.(A)/(2) cos.(B)/(2) cos.(C)/(2)`
C. `2r^(2) sin (A -B) sin (B -C) sin(C -A)`
D. none of these

Answer

B

Explanation

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)`