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 ?
Please help.
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.
Remark: You probably do not need explanation of the ratios. It comes down to the fact that the ratio of the areas of triangles of the same height is equal to the ratio of their bases.