1.

Let the incircle with center I of DeltaABC touch sides BC, CA, AB at D,E and F, respectively. Let a circle is drawn touching ID, IF and incircle of DeltaABC having radius r_(2). Similarly r_(1) and r_(3) are defined. Prove that (r_(1))/(r -r_(1)) .(r_(2))/(r-r_(2)) .(r_(3))/(r-r_(3)) = (a + b+c)/(8R)

Answer»

SOLUTION :
From the figure, in `Delta IQP`
`cos.(B)/(2) = (PQ)/(IP)`
Since `PQ = r_(2)`
and `IP = r - r_(2)` (as two circles TOUCHING internally), we have
`cos.(B)/(2) = (r_(2))/(r-r_(2))`
SIMILARLY, for other such circles, `cos.(A)/(2) = (r_(1))/(r -r_(1)) and cos.(C)/(2) = (r_(3))/(r -r_(3))`
`(r_(1))/(r -r_(1)) .(r_(2))/(r -r_(2)) .(r_(3))/(r -r_(3)) = cos.(A)/(2) cos.(B)/(2) cos.(C)/(2)`
`= (1)/(4) (sin A+ sin B + sin C)`
`= (a + b + c)/(8R)`


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