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AD, BE, CF are internal angular bisectors of Delta ABC and I is the incentre. If a(b+c)sec.(4)/(2)ID+b(a+c)sec.(B)/(2)IE+c(a+b)sec.(C )/(2)IF=kabc, then the value of k is |
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Answer» 1 `(ID)/(AD)=(ID)/((2bc)/(b+c)cos.(A)/(2))=(a(b+c)SEC.(A)/(2).ID)/(2abc)` Now `(AI)/(ID)=(AB)/(BD)=(c )/((ac)/(b+c))=(b+c)/(a)` `therefore (ID)/(AD)=(ID)/(AI+ID)=(a)/(a+b+c)` `therefore(ID)/(AD)+(IE)/(BE)+(IF)/(CF)=(a+b+c)/(a+b+c)=1` `therefore a(b+c)sec.(A)/(2)ID+b(a+c)sec.(B)/(2)IE + c(a+b)sec.(C )/(2)IF` = 2abc `therefore K = 2` |
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