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Let $ABCD$ be a square and $M$ a point on $BC$.
$DM \cap AB=\{E\}$ and $AM \cap CD=\{F\}$.

Prove that $\displaystyle \frac{BE}{AE}+\frac{CF}{FD}=1$.

thanks:)

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    Hint: Use the line through $M$, parallel to $AB$ and apply the Intercept theorem 4 times.2012-09-02

2 Answers 2

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From isometry of $\triangle BEM$ and $\triangle AED$:

$\frac{BE}{AE}=\frac{BM}{AD}=\frac{BC-MC}{AD}=1-\frac{MC}{AD}$ since $AD=BC$

From isometry of $\triangle MFC$ and $\triangle AFD$: $\frac{MC}{AD}=\frac{CF}{FD}$ and the result follows. Now your task is to draw the picture ;)

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    sure, thanks. couldn't read my own scribblings2012-09-02
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Let $M'$ be the point on $AD$ such that $AM'=BM$. By applying the intercept theorem twice we get

$\frac{BE}{AE}=\frac{ME}{DE}=\frac{AM'}{AD}$

and similarly

$\frac{CF}{FD}=\frac{MF}{AF}=\frac{DM'}{AD}$

Together

$\frac{BE}{AE}+\frac{CF}{FD}=\frac{AM'+DM'}{AD}=1$