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Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question Area of DeltaDEF is |
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Answer» `2r^(2) sin(2A) sin(2B) sin(2C)` 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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