Incenters Are Important

@Deepansh Jindal – I will give you a short insight of the proof-

Consider the same triangle as I used in the generalized formula and my solution. We know that angle bisectors divide the side on which fall into the same ratio of other two sides (see solution).

So,

$AP=\frac{bc}{a+b}$ , $PB=\frac{ac}{a+b}$ , $BQ=\frac{ac}{b+c}$ , $QC=\frac{ab}{b+c}$ , $CR=\frac{ab}{a+c}$ and $RA=\frac{bc}{a+c}$ .

Also,

$\sin \angle A = \frac{2}{bc}\times ar(\bigtriangleup ABC$ ), $\sin \angle B = \frac{2}{ca}\times ar(\bigtriangleup ABC$ ) and $\sin \angle C = \frac{2}{ab}\times ar(\bigtriangleup ABC$ ).

Now using this information to calculate area of $\bigtriangleup APR$ , $\bigtriangleup BQP$ and $\bigtriangleup CRQ$ .

$ar(\bigtriangleup APR)=\frac{bc}{(a+b)(b+c)}\times ar(\bigtriangleup ABC)$

$ar(\bigtriangleup BQP)=\frac{ca}{(b+c)(a+b)}\times ar(\bigtriangleup ABC)$

$ar(\bigtriangleup CRQ)=\frac{ab}{(c+a)(b+c)}\times ar(\bigtriangleup ABC)$

Now I leave the rest to you, you just need to subtract these areas with the area of $\bigtriangleup ABC$ (as done in the solution), after lots of algebra and simplification you will get the answer.

@Deepansh Jindal – @Deepansh Jindal I have tried my best to explain the proof to you.