In triangle ABC a point X is taken on AC and a point Y is taken on BC if AY and BX meet at O This question is from pre collage mathematics.
The question goes on like this:

In $\triangle ABC$ a point $X$ is taken on $\overline{AC}$ and a point $Y$ is taken on $\overline{BC}$. If $\overline{AY}$ and $\overline{BX}$ meet at $O$, find the area of $\triangle CXY$ if the areas of triangles $OXA$, $OAB$ and $OBY$ are $x,y,z$ respectively.

My answer is:
$$\frac{xz(x+y)(y+z)}{y(y^2-zx)}$$

$\frac{A(ABY)}{A(ACY)}=\frac{A(BXY)}{A(CXY)}=\frac{BY}{CY}$
$\frac{y+z}{x+A(OXY)+A(CXY)}=\frac{z+A(OXY)}{A(CXY)}$
$A(CXY)\times(y+z)=(x+A(OXY)+A(CXY))\times(z+A(OXY))$
$y\times A(CXY)=xz+x\times A(OXY)+z\times A(OXY)+(A(OXY))^2+A(CXY)\times A(OXY)$
$A(CXY)=\frac{xz+x\times A(OXY)+z\times A(OXY)+(A(OXY))^2}{y- A(OXY)}=\frac{xz+\frac{x^2z}{y}+\frac{xz^2}{y}+\frac{x^2z^2}{y^2}}{y-\frac{xz}{y}}=\frac{y^2+xy+yz+xz}{y(y^2-xz)}=\frac{xz(x+y)(y+z)}{y(y^2-xz)}$
and using this
$\frac{A(ABO)}{A(AXO)}=\frac{A(OBY)}{A(OXY)}=\frac{BO}{XO}$
that is
$\frac{y}{x}=\frac{z}{A(OXY)}$
then
${A(OXY)}=\frac{xz}{y}$
