Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular.
(B. Kukushkin)
By perpendicularity lemma, it is sufficient to prove that $BF^2-BD^2=EF^2-ED^2$ which is true by Pythagorean theorem for right triangles $\triangle BAF$ and $\triangle BCD$
Construct circles with centres $D$ and $F$ and with radii $FA$ and $DC$ respectively. Then the point $B$ lies on the radical axis of the circles, since $BA=BC$ and $BA$, $BC$ are tangent to the circles. Obviously $E$ lies on both circles. This means that the line $BE$ is the radical axis of the two circles, meaning it is perpendicular to the line which connects centres which is $DF$ $\blacksquare$
