# a geometry problem. $\frac{BE}{AE}+\frac{CF}{FD}=1$

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. – Simon Markett Sep 2 '12 at 17:34

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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AEB is not a triangle – Iuli Sep 2 '12 at 17:42
I think you mean AED. – Jair Taylor Sep 2 '12 at 17:43
sure, thanks. couldn't read my own scribblings – Valentin Sep 2 '12 at 17:45

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

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

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