Internal bisectors of `DeltaABC` meet the circumcircle at point D, E, and F
Area of `DeltaDEF` is

Options

A. `2R^(2) cos^(2) ((A)/(2)) cos^(2)((B)/(2)) cos^(2) ((C)/(2))`
B. `2R^(2) sin ((A)/(2)) sin ((B)/(2)) sin((C)/(2))`
C. `2R^(2) sin^(2) ((A)/(2)) sin^(2) ((B)/(2) sin^(2) ((C)/(2))`
D. `2R^(2) cos((A)/(2)) cos((B)/(2)) cos((C)/(2))`

Answer

D

Explanation

Area of `DeltaABC = (abc)/(4R) = 2R^(2) sin A sin B sin C`
`angleADE = angleABE = (B)/(2)`
Similarly, `angleFDA = angleFCA = (C)/(2)`
`rArr angleFDE = (B+C)/(2)`
and `angleDEF = (A+C)/(2) and angleDFE = (A +B)/(2)`

In `DeltaDEF`, by sine rule, `(EF)/(sin (angleFDE)) = 2R`
`rArr EF = 2R cos ((A)/(2))`
Then, area of `DeltaDEF`
`= 2R^(2) sin ((A+B)/(2)) sin ((B+C)/(2)) sin ((A+C)/(2))`
`=2R^(2) cos ((A)/(2)) cos ((B)/(2)) cos ((C)/(2))`
Now `(Delta_(ABC))/(Delta_(DEF)) = (2R^(2) sin A sin B sin C)/(2R^(2) cos ((A)/(2)) cos((B)/(2)) cos((C)/(2)))`
`= 8sin ((A)/(2)) sin ((B)/(2)) sin ((C)/(2)) le 1`
`rArr Delta_(ABC) le Delta_(DEF)`