Ratio of parts of an intersected segment in a rectangle 
In rectangle $ABCD$, points $E$ and $F$ lie on sides $BC$ and $CD$ respectively. Point $F$ is the midpoint of $CD$ and $BE=\frac13BC$. Segments $AC$ and $FE$ intersect at point $P$. What is the ratio of $AP$ to $PC$? Express your answer as a common fraction.


I'm not sure where to start, do I need to use the fact that $FCE$ is a right triangle, or just the ratios, any help would be appreciated.

$A = (0,0)$
$B = (x,0)$
$C = (x,-y)$
$D = (0,-y)$
$E = (x,-y/3)$
$F = (x/2,-y)$
$P = (tx,-ty)$
$P = \lambda  E + (1-\lambda)F$
$tx = \lambda  x + (1-\lambda)x/2 = \lambda  x + x/2 - \lambda x/2 = (\lambda + 1/2 - \lambda/2) x $
$t = \frac{(\lambda + 1)}{2}$
$-\frac{(\lambda + 1)}{2}y = \lambda (-y/3) + (1-\lambda)(-y) = \lambda (-y/3) + -y + \lambda y = (\frac{2 \lambda}{3} - 1) y$
$-\frac{(\lambda + 1)}{2} = (\frac{2 \lambda}{3} - 1) \text{ iff }  \lambda = \frac{3}{7} \text{ iff } t = \frac{5}{7}  \text{ iff } 1-t = \frac{2}{7}$
$\text{Length of } AP = \sqrt{ (\frac{5}{7}x)^2 + (\frac{5}{7}y)^2}$
$\text{Length of } BC = \sqrt{ (\frac{2}{7}x)^2 + (\frac{2}{7}y)^2}$
ANS: $\frac{Len(AP)}{Len(PC)} = \frac {\frac{5}{7}}  {\frac{2}{7}} = \frac {5}{2} $
Note: From $[\lambda = \frac{3}{7}]$ and $[1 - \lambda = \frac{4}{7}]$  we also get $\frac{Len(FP)}{Len(PE)} = \frac{3}{4}$ 'free of charge'.
