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AD, BE, CF are internal angular bisector...

AD, BE, CF are internal angular bisectors of `Delta ABC` and I is the incentre. If `a(b+c)sec.(A)/(2)ID+b(a+c)sec.(B)/(2)IE+c(a+b)sec.(C )/(2)IF=kabc`, then the value of k is (a) 1 (b) 2 (c) 3 (d) 4

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
B

`(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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