The area of a triangle determined by the bisectors. How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture: 
 
$AQ$ is the bisector of the angle $\angle BAC$, $BR$ -bisector for $\angle ABC$ and $CP$ -bisector for $\angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$. 
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything. 
Thanks :) 

We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, \quad (I)$$
Using the angle bisector theorem we get:
$$BP=\frac{ac}{a+b},\quad (1)$$
$$BR=\frac{ac}{b+c}, \quad (2)$$
$$CR=\frac{ab}{b+c},\quad (3)$$
$$CQ=\frac{ab}{a+c},\quad (4)$$
$$AQ=\frac{bc}{a+c},\quad (5)$$
and
$$AP=\frac{bc}{a+b}. \quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=\frac{1}{2}ab\sin\gamma, \quad (7)$$
$$A_{PBR}=\frac{1}{2}BP\cdot BR\sin\beta, \quad (8)$$
$$A_{RCQ}=\frac{1}{2}CR\cdot CQ\sin\gamma, \quad (9)$$
and
$$A_{QAP}=\frac{1}{2}AQ\cdot AP\sin\alpha. \quad (10)$$
Let $R$ be the circumradius, we  know that:
$$\sin \alpha = \frac{a}{2R}, \quad (11)$$
$$\sin \beta = \frac{b}{2R}, \quad (12)$$
$$\sin \gamma = \frac{c}{2R}, \quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=\frac{1}{2} \cdot \frac{abc}{2R}-\frac{1}{2} \frac{a^2c^2b}{(a+b)(b+c)2R}-\frac{1}{2} \cdot \frac{a^2b^2c}{(b+c)(a+c)2R}-\frac{1}{2} \cdot \frac{b^2c^2a}{(a+b)(a+c)2R}, \Rightarrow$$
$$A_{PQR}=\frac{abc}{4R}[1-\frac{ac}{(a+b)(b+c)}-\frac{ab}{(b+c)(a+c)}-\frac{bc}{(a+b)(a+c)}], \Rightarrow$$
$$A_{PQR}=\frac{abc}{2R}[\frac{abc}{(a+b)(b+c)(a+c)}], \Rightarrow$$
$$A_{PQR}=A_{ABC}[\frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
