Area of the quadrilateral within a triangle Given the area of tringles $BEF=X,BFC=Y$ and $FDC=Z$, Can we find the area of the quadrilateral $AEFD$ in terms of $X,Y,Z$?

 A: Let $U = AEF$ and $V = ADF$. Note that $AEFD = U + V$.
$BCE\frac{AB}{BE} = ABC \Rightarrow (X+Y)\frac{AB}{BE} = X + Y + Z + U + V$
$BEF\frac{AB}{BE} = ABF \Rightarrow X\frac{AB}{BE} = X + U$
Dividing this two equations we get:
$\frac{X+Y}{X} = \frac{X + Y + Z + U + V}{X + U} \Rightarrow (X + U)\frac{X+Y}{X} = X + Y + Z + U + V \Rightarrow X + Y + U + U\frac{Y}{X} = X + Y + U + V \Rightarrow \boxed{U\frac{Y}{X} = Z + V}$
Similarly:
$BCD\frac{AC}{CD} = ABC \Rightarrow (Y + Z)\frac{AC}{CD} = X + Y + Z + U + V$
$CDF\frac{AC}{CD} = ACF \Rightarrow Z\frac{AC}{CD} = Z + V$
Dividing this two equations we get:
$\frac{Y + Z}{Z} = \frac{X + Y + Z + U + V}{Z + V} \Rightarrow (Z + V)\frac{Y + Z}{Z} = X + Y+ Z + U + V \Rightarrow Y + Z + V + V\frac{Y}{Z} = X + Y + Z + U + V\Rightarrow \boxed{V\frac{Y}{Z} = X + U}$
We have a nice system of two equations with two unknowns $U$ and $V$.
This can be easily solved:
$$U = \frac{XYZ + X^2Z}{Y^2-XZ}$$
$$V = \frac{XYZ + XZ^2}{Y^2-XZ}$$
The area of the quadrilateral is $U + V$:
$$AEFD = U + V = \frac{2XYZ + X^2Z + XZ^2}{Y^2-XZ}$$
