0
$\begingroup$

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:)

  • 0
    Hint: Use the line through $M$, parallel to $AB$ and apply the Intercept theorem 4 times.2012-09-02

2 Answers 2

2

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 ;)

  • 0
    AEB is not a triangle2012-09-02
  • 0
    I think you mean AED.2012-09-02
  • 0
    sure, thanks. couldn't read my own scribblings2012-09-02
1

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

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