Let the incircle of a `Delta ABC` touches sides BC, CA and AB at D,E and F, respectively. Let area of `Delta ABC` be `Delta` and thatof DEF be `Delta'`. If a, b and c are side of `Detla` `ABC`, then the value of `abc(a+b+c)(Delta')/(Delta^(3))` is (a) 1 (b) 2 (c) 3 (d) 4

Options

A. 1
B. 2
C. 3
D. 4

Answer

D

Explanation

Clearly points C, D, I and E are concyclic
`therefore angle EID = pi - C`
`therefore` In `Delta DEF, angle DFE =(pi-C)/(2)`
Similarly `angle EDF =(pi-A)/(2)` and `angle FED = (pi-B)/(2)`
`Delta' =2r^(2)sin((pi-A)/(2))sin((pi-B)/(2))sin((pi-C)/(2))`
`=2r^(2)cos((A)/(2))cos((B)/(2))cos((C )/(2))`
`therefore abc (a+b+c)(Delta')/(Delta^(3))`
`=(abc(a+b+c))/(Delta^(3))2r^(2)cos.(A)/(2)cos.(B)/(2)cos.(C )/(2)`
`=(abc(2s))/(Delta^(3))2r^(2)cos.(A)/(2)cos.(B)/(2)cos.(C )/(2)`
`=(abc(2s))/(Delta^(3))2r^(2)sqrt((s(s-a))/(bc))sqrt((s(s-b))/(ab))sqrt((s(s-c))/(ac))`
`=(4abcs)/(Delta^(3))r^(2)(s)/(abc)sqrt(s(s-a)(s-b)(s-c))`
`=(4s^(2))/(Delta^(2))r^(2)`
= 4.