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Question

Let ABC be a triangle with incentre I and r. Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA and AB respectively, If r1,r2, andr3 are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, prove that
r1rr1+r2rr2+r3rr3=r1r2r3(rr1)(rr2)(rr3)

A
True
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B
False
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Solution

The correct option is A True
Let MN=r3=MP=MQ,ID=r
IP=rr3
Clearly IP and IQ are tangents to circle with centre M.
IM must be the bisector of PIQ
PIM=QIM=θ1
Also from ΔIPM,tanθ1=r3rr3=MPIP


Here DI=r
Similarly, in other quadrilaterals, we get
tanθ2=r2rr2 and tanθ3=r1rr1
Also 2θ1+2θ2+2θ3=2πθ1+θ2+θ3=π
tanθ1+tanθ2+tanθ3=tanθ1.tanθ2.tanθ3
r1rr1+r2rr2+r3rr3=r1r2r3(rr1)(rr2)(rr3)

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