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 ?