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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 angleDEF is equal to |
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Answer» `(pi -B)/(2)` 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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