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: Let's see the figure below:

From the similarity of the triangles $FCE$ and $BEQ$ we get that $a'=\frac a4$.
Then from the similarity of the triangles $FCP$ and $PAQ$ we get that
$$\frac xy=\frac{a+\frac a4}{\frac a2}=\frac52.$$
A: Let the area of $ABCD$ be 12 units. From there we can see clearly see that


*

*$AFD$'s area is 3 units

*$ABE$'s is 2 units

*$CFE$'s is 2 units by comparison with $AFD$, which has the same base length and $\frac32$ times the height

*$AEF$'s is 5 units by subtraction


Now because the two triangles $AEC$ and $ACF$ share a common base, and $EF$ cuts them in a 5:2 ratio, $\frac{AP}{PC}=\frac52$.
A: 
The lines $y=\frac{2}{3}x$ and $y=\frac{3}{2}-\frac{x}{2}$ intersect at $x=\frac{9}{7}$, hence by Thales' theorem
$$ \frac{AP}{PC} =  \frac{3+\frac{9}{7}}{3-\frac{9}{7}}=\color{red}{\frac{5}{2}}. $$
You are free to assume that $ABCD$ is a $6\times 3$ rectangle since linear maps preserve the ratios of segments cut on the same line.
A: Using vector method, let $D$ be the origin. 
Let $P$ divide $AC$ in ratio $k:1$, so $\vec{DP}=\frac{k\vec{DC}+\vec{DA}}{k+1}$
Also, let $P$ divide $EF$ in ratio $m:1$, we again get $\vec{DP}=\frac{m\vec{DF}+\vec{DE}}{m+1}$
Can you now proceed?
A: $A = (0,0)$
$B = (x,0)$
$C = (x,-y)$
$D = (0,-y)$
$E = (x,-y/3)$
$F = (x/2,-y)$
$P = (tx,-ty)$
$P = \lambda  E + (1-\lambda)F$
$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'.
