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 $r_1,r_2,and{r}_3$ are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, prove that
$\frac{r_1}{r-r_1}+\frac{r_2}{r-r_2}+\frac{r_3}{r-r_3}=\frac{r_1{r}_2{r}_3}{(r-r_1)(r-r_2)(r-r_3)}$

• A. True
• B. False

Answer 1

The correct option is A True

Let $MN=r_3=MP=MQ,ID=r$
$\Rightarrow IP=r-r_3$
Clearly IP and IQ are tangents to circle with centre M.
$\therefore$ IM must be the $\angle$ bisector of $\angle PIQ$
$\therefore$ $\angle PIM=\angle QIM={\theta }_1$
Also from $\Delta IPM,tan{\theta }_1=\frac{r_3}{r-r_3}=\frac{MP}{IP}$


Here DI=r
Similarly, in other quadrilaterals, we get
$tan{\theta }_2=\frac{r_2}{r-r_2}$ and $tan{\theta }_3=\frac{r_1}{r-r_1}$
Also $2{\theta }_1+2{\theta }_2+2{\theta }_3=2\pi \Rightarrow {\theta }_1+{\theta }_2+{\theta }_3=\pi$
$\Rightarrow tan{\theta }_1+tan{\theta }_2+tan{\theta }_3=tan{\theta }_1.tan{\theta }_2.tan{\theta }_3$
$\Rightarrow \frac{r_1}{r-r_1}+\frac{r_2}{r-r_2}+\frac{r_3}{r-r_3}=\frac{r_1{r}_2{r}_3}{(r-r_1)(r-r_2)(r-r_3)}$