Prove that $\dfrac{r_1}{r-r_1}+\dfrac{r_2}{r-r_2}+\dfrac{r_3}{r-r_3}=\dfrac{r_1r_2r_3}{(r-r_1)(r-r_2)(r-r_3)}$ for triangle radii Let ABC be a triangle with incentre  $I$  and  inradius  $r$ . Let  $D,E,F$  be the feet of the perpendiculars from  $I$  to the sides $BC,CA,AB$ respectively.If $r_1 ,r_2,r_3$ are the radii of circles inscribed in the quadrilaterals  $AFIE,BDIF,CEID$ respectively, prove that
$$\dfrac{r_1}{r-r_1}+\dfrac{r_2}{r-r_2}+\dfrac{r_3}{r-r_3}=\dfrac{r_1r_2r_3}{(r-r_1)(r-r_2)(r-r_3)}$$
My attempt is as follows:-

$I'$ is the incenter of the quadilateral AFIE (Fig $2$)
In $\triangle AGI'$, $\dfrac{GI'}{AG}=\tan\dfrac{A}{2}$ where $GI'=r_1$ and $AG=x$
In $\triangle$ $AFI$, $\dfrac{FI}{AF}=\tan\dfrac{A}{2}$ where $FI=r$ and $AF=s-a$
$$\dfrac{r_1}{x}=\dfrac{r}{s-a}$$
All I need here is the value of $x$
In Fig $2$
$$AG=x\tag{1}$$
$$GF=AF-AG$$
$$GF=s-a-x\tag{2}$$
$$FM=s-a-x$$
$$MI=FI-FM$$
$$MI=r-s+a+x$$
$$IN=r-s+a+x$$
$$NE=r-\left(r-s+a+x\right)=s-a-x$$
$$EH=s-a-x$$
$$HA=EA-EH$$
$$HA=s-a-\left(s-a-x\right)$$
$$HA=x$$
Initially I thought that $HA$ would come something other than $x$ and I will equate $AG$ and $HA$, but unfortunately I got $HA$ also as $x$.
Then I observed that quadilateral $AFIE$ is cyclic as $\angle AFI+\angle AEI=\pi$. But I didn't get how to take advantage of it.
I was just able to say $\angle FIE=\pi-A$
Please help me in this.
 A: Continue with what you have $\dfrac{r_1}{x}=\dfrac{r}{s-a}$ and recognize FG $= s-a-x=r_1$ in your diagram to get
$$\frac{s-a}r=\frac{r_1}{r-r_1}$$
Then, use the triangle area $K=\frac12(a+b+c)r$ to express $$\dfrac{r_1}{r-r_1}+\dfrac{r_2}{r-r_2}+\dfrac{r_3}{r-r_3}
=\frac{s-a}r + \frac{s-b}r +\frac{s-c}r = \frac sr=\frac K{r^2}$$
$$\dfrac{r_1r_2r_3}{(r-r_1)(r-r_2)(r-r_3)} = \frac1{r^3}(s-a)(s-b)(s-c)=\frac K{r^2}$$
Thus, LHS = RHS.
A: In any polygon with an inscribed circumference and area $A$, its inradius $r$ and its perimeter $p$ will satisfy $$pr=2A.$$ The proof is identical to that of the triangular case. We can therefore calculate $$r_1=\frac{(s-a)\cdot r}{(s-a)+r},$$ $$r_2=\frac{(s-b)\cdot r}{(s-b)+r},$$ $$r_3=\frac{(s-c)\cdot r}{(s-c)+r}.$$ If we substitute these values into the equation you want to prove, and simplify, we’ll get the equivalent statements $$\frac{s-a}{r}+\frac{s-b}{r}+\frac{s-c}{r}=\frac{\left(s-a\right)\cdot\left(s-b\right)\cdot\left(s-c\right)}{r^3}\Leftrightarrow$$ $$sr=\sqrt{s\cdot\left(s-a\right)\cdot\left(s-b\right)\cdot\left(s-c\right)}.$$ But these are both equal to the area of $\bigtriangleup ABC$, and we’re done. $\blacksquare$
