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In rectangle $ABCD$ points $E$ and $F$ lie on $BC$ and $AD$ respectively so that $AF=2FD$ and $3BE=4EC$. Call the intersection of $AE$ and $BF$ as $X$ and the intersection of $AE$ and $BD$ as $Y$.Calculate $\frac{XY}{AE}$

I think Thales theorem and similar triangles should help but don't know how to use them...

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    Is the answer $\frac{14}{143}$? – 2017-01-21
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    I don't know the answer! – 2017-01-21

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From $F$ and $D$ draw two lines parallel to $AE$. These two lines intersect the line of $BC$ at $G$ and $H$ respectively. It is evident that $AE=FG=DH$ and $AF=EG$ and also $FD=GH$. Let $K$ be the point of intersection of $FG$ and $BD$. Then we can write: $$\begin{align}\frac{FK}{AY}&=\frac{DF}{AD}=\frac 13\\ \frac{XY}{FK}&=\frac{BY}{BK}=\frac{BE}{BG}=\frac{4/7}{4/7+2/3}=\frac 6{13}\\ \implies XY&=\frac 6{13}FK=\frac 6{13}\times\frac 13 AY \end{align}$$ On the other hand $$\frac{YE}{AE}=\frac{YE}{DH}=\frac{BE}{BH}=\frac{4/7}{1+4/7}=\frac 4{11}\\ \implies\frac{AY}{AE}=1-\frac{YE}{AE}=\frac 7{11}$$ In conclusion $$XY=\frac 2{13}AY=\frac 2{13}\times\frac 7{11}AE\implies\frac{XY}{AE}=\frac{14}{143}$$

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    But I think there must be a simpler method not requiring to draw any extra segments... – 2017-01-25
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    @HamidRezaEbrahimi There is a workaround to avoid $DH$. But I think $FG$ is inevitable. Nonetheless, why should anyone want to avoid drawing additional lines if they help us to simplify and better understand the problem? – 2017-01-25
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    I think if possible and not necessary , we should avoid complicating the picture – 2017-01-26