# A problem related to area of triangles.

I'm solving some practice problems to prepare for a competitive exam . Here is one which I'm trying to do for some time but still haven't found a solution to :

" In $ΔABC$ , $E$ and $F$ are such that $A-F-B$ and $A-E-C$ . Segments $BE$ and $CF$ intersect at $P$. Area of $ΔPBC = 10$, Area of $ΔPEC = 4$ and area of $ΔPFB = 8$. Find the area of quadrilateral $AFPE$ . "

Here is the drawing I made :

Now for finding the area of quadrilateral $AFPE$ I'll have to find the area of $ΔABC$ and then subtract the areas of three triangles. But I haven't been able to find the area of $ΔABC$ .

Since triangles $ΔPEC$ and $ΔPBC$ have equal heights corresponding to their bases $PE$ and $PB$ , I get $PE/PB=$ratio of their areas $= 4/10=2/5$ and similarly $FP/PC=4/5$ but what now ?

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Draw the line $AP$. Let $\triangle AFP$ have area $x$, and let $\triangle AEP$ have area $y$.
Then $\dfrac{8+x}{y}=\dfrac{10}{4}$ and $\dfrac{y+4}{x}=\dfrac{10}{8}$. We obtain two linear equations in $x$ and $y$. Solve.
I get the same thing, which I prefer to write as $\frac{256}{17}$. – André Nicolas Sep 10 '13 at 18:08