Are you sure you didn't make a typo? Right now, $E$ is just the midpoint of $CD$ so by symmetry $AE \perp BE$ just means that $\angle EBA=45^\circ$ and hence $AB=2BC$ so that $E=F$. But then certainly $BF>AB$...

Tintarn wrote: Are you sure you didn't make a typo? Right now, $E$ is just the midpoint of $CD$ so by symmetry $AE \perp BE$ just means that $\angle EBA=45^\circ$ and hence $AB=2BC$ so that $E=F$. But then certainly $BF>AB$... Thanks I did change it.

Let $X$ be the intersection of $BE$ and $AC$. From it we have right-angled triangles: $\triangle ABC$, $\triangle BXC$, and $\triangle BCE$ $\Rightarrow$ $\angle BAC$ = $90$ − $\angle BCA$ = $90$ − $\angle BCX$ = $\angle CBX$ = $\angle CBE$ Since $\angle ACB=\angle CBA$ = $90$ $\Rightarrow$ $\triangle ECB$ $\backsim$ $\triangle CBA$. From the last relation we get: $\frac{CE}{BC}$= $\frac{BC}{AB}$ $\Rightarrow$ $\frac{AB}{2BC}$= $\frac{BC}{AB}$ $\Rightarrow$ $(AB)^2$ = $2(BC)^2$ $\Rightarrow$ $AB$ $\sqrt 2$$BC$. And last, by using pythagorean theorem for $\triangle BCF$ we get : $BF^2$ = $BC^2$ + $CF^2$ = $2BC^2$ $\Rightarrow$ $BF$ = $\sqrt 2$$BC$ = $AB$.

I may have gone through a longer route but still $ \angle BAC=90^{\circ}- \angle BCA = \angle CBG$ $AB||CD \Rightarrow \angle BAC = \angle ACE \Rightarrow \angle CBG = \angle ACE \Rightarrow \angle BCA = \angle CEB$ $\Rightarrow \bigtriangleup ABC \sim \bigtriangleup BCE$ $\frac{BC}{AB}= \frac{CE}{BC} \Rightarrow \frac{BC}{AB}= \frac{AB}{2BC}$ $\Rightarrow AB^2=2BC^2 \Rightarrow AB= \sqrt{2} BC$ Using the pythagorean theorem at $\bigtriangleup BCF$, we get: $BF^2=BC^2+CF^2=2BC^2 \Rightarrow BF= \sqrt{2}BC=AB$ $\blacksquare$

λύση αργότερα,