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Internal bisectors of DeltaABC meet the circumcircle at point D, E, and F Length of side eF is |
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Answer» `2R cos.(A)/(2)` `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)` |
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