In $\triangle ABC$, I is the incenter. Area of $\triangle IBC = 28$, area of $\triangle ICA= 30$ and area of $\triangle IAB = 26$. Find $AC^2 − AB^2$ In $\triangle ABC$, I is the incenter. Area of $\triangle IBC = 28$, area of $\triangle ICA = 30$ and area of $\triangle IAB = 26$. Find $AC^2 − AB^2$.
Here is a sketch that I drew:


From the given areas ,
$r*AC=60$
$r*AB=52$
$r*BC=56$
So that I've three equations with 4 unknowns . 
Also,
$A=rs$
$2A=168=r(AB+BC+AC)$
But this can't be used as it can be derived from the other three equations.
So, what should I do? Any hints are apreciated.
(This is not class-homework , I'm solving sample questions for a competitive exam )  
 A: we note $AC=b,BC=a,AB=c$, we will find $b^2-c^2$. We have
$$ar=56,br=60,cr=52.$$
then we have the proportion of $a$ and $b,c$:
$$b=\frac{60}{56}a,c=\frac{52}{56}a$$
Then we have the equality:
$$a=\frac{56}{112}(b+c)=\frac{1}{2}(b+c)$$
$$a=\frac{56}{8}(b-c)=7(b-c)$$
Now we should recall the formula of triangle's area:
$S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}$
Then we have
$$84=\sqrt{\frac{a+b+c}{2}\cdot\frac{b+c-a}{2}\cdot\frac{a+c-b}{2}\cdot\frac{a+b-c}{2}}\\=\frac{1}{4}\sqrt{[(b+c)^2-a^2][a^2-(b-c)^2]}\\=\frac{1}{4}\sqrt{[(b+c)^2-\frac{1}{4}(b+c)^2][49(b-c)^2-(b-c)^2]}\\=\frac{1}{4}\sqrt{\frac{3}{4}(b+c)^2\cdot48(b-c)^2}\\\frac{3}{2}|b^2-c^2|\\=\frac{3}{2}(b^2-c^2)$$(because $b>c$).
then we have what we want.
A: A remark on the solvability of the question: you are given $ra=56,rb=60,rc=52$ (with the usual notations $r$ the inradius, $a=BC$, etc). Therefore you can deduce that the triangle $ABC$ is similar to the triangle $\Delta=(56,60,52)$ with scaling ratio $r$. 
You find the area of $ABC$ by summing up the areas of the three triangles and use the fact that $r^2 Area(ABC)=Area(\Delta)$ (you can calculate $Area(\Delta)$). This gives you $r$, and therefore you can find $a,b,c$.
A: Note that $b=\frac{15a}{14}$, $c=\frac{13a}{14}$. Thus, $s=\frac{14a+13a+15a}{2\cdot14}=\frac{3a}2$. So by Heron's formula, $$84=\sqrt{\frac{3a^2}4\cdot \frac{3a}7\cdot \frac{4a}7}=\frac{6a^2}{14}$$
Thus, $a^2=14^2$, or $a=14$. Now you can go on from here.
