Let $ABCD$ be a rectangle with $AB \ne BC$ and the center the point $O$. Perpendicular from $O$ on $BD$ intersects lines $AB$ and $BC$ in points $E$ and $F$ respectively. Points $M$ and $N$ are midpoints of segments $[CD]$ and $[AD]$ respectively. Prove that $FM \perp EN$ .
Problem
Source: Danube 2013 Junior P4
Tags: geometry, rectangle, perpendicular
achen29
07.09.2018 11:52
Isn't O like on BD?
Tintarn
07.09.2018 12:02
achen29 wrote: Isn't O like on BD? Sure. So what? We can still draw the perpendicular line?
Tintarn
07.09.2018 12:23
So it clearly suffices to prove that triangles $AEN$ and $CFM$ are similar (then since $AE \perp CF$ this will imply $EN \perp FM$ as desired).
If we let $AD=a, AB=b, BD=d$ it will suffice to prove that $\frac{AE}{CF}=\frac{a}{b}$. But we can simply compute these lengths.
By similarity of $BFO$ and $DBA$ we find that $BF=\frac{d^2}{2a}$ and similarly $BE=\frac{d^2}{2b}$. Hence
\[\frac{AE}{CF}=\frac{b-BE}{BF-a}=\frac{b-\frac{d^2}{2b}}{\frac{d^2}{2a}-a}=\frac{a}{b} \cdot \frac{2b^2-d^2}{d^2-2a^2}=\frac{a}{b}\]using that $d^2=a^2+b^2$. Done.
sunken rock
08.09.2018 10:07
Actually it suffices to prove $\triangle DAE\sim\triangle DCF$, thus $DF\perp DE$ and thence $FM\bot EN$ as homologous medians. But this is very easy, since $AEOD$ and $DOCF$ are cyclic, making $\angle ADE=\angle AOE=\angle COF=\angle CDF$. Other solution is to take $P$ midpoint of $AB$ and to prove $\triangle PEN\sim\triangle OMF$. Best regards, sunken rock
