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