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Solution of Triangle

Let ABC be a ...

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 $r_{1}, r_{2}, a n d r_{3}$ are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, prove that
$\frac{r_{1}}{r - r_{1}} + \frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}$

True

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False

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Solution

The correct option is A True
Let $M N = r_{3} = M P = M Q, I D = r$
$\Rightarrow I P = r - r_{3}$
Clearly IP and IQ are tangents to circle with centre M.
$∴$ IM must be the $∠$ bisector of $∠ P I Q$
$∴$ $∠ P I M = ∠ Q I M = θ_{1}$
Also from $Δ I P M, t a n θ_{1} = \frac{r_{3}}{r - r_{3}} = \frac{M P}{I P}$

Here DI=r
Similarly, in other quadrilaterals, we get
$t a n θ_{2} = \frac{r_{2}}{r - r_{2}}$ and $t a n θ_{3} = \frac{r_{1}}{r - r_{1}}$
Also $2 θ_{1} + 2 θ_{2} + 2 θ_{3} = 2 π \Rightarrow θ_{1} + θ_{2} + θ_{3} = π$
$\Rightarrow t a n θ_{1} + t a n θ_{2} + t a n θ_{3} = t a n θ_{1} . t a n θ_{2} . t a n θ_{3}$
$\Rightarrow \frac{r_{1}}{r - r_{1}} + \frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}$

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\frac{r_{1}}{r - r_{1}} + \frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}

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