Estonian Mathematical Olympiad

As $AB$ is a diameter of $c$, $\angle AFB=90^{\circ }$ (Fig. 27) and $BF$ is an altitude of triangle $ABC$. From $AB=BC$, $BF$ is a median and $F$ is the midpoint of $AC$. By symmetry, point $K$ lies on $BF$ and $\angle FKE=90^{\circ }$. Notice that $\angle BAC=\angle CAD=\angle DAE$, as $BAC,CAD$, and $DAE$ are inscribed angles that subtend to equal arcs of the circumcircle of the regular pentagon $ABCDE$. Points $A,B,F$ and $G$ lie on circle $c$ in this order, thus $\angle ABF=180^{\circ }-\angle AGF=\angle AGH$. Hence, $ABF$ and $AGH$ are similar from two angles, implying $\angle AHG=\angle AFB=90^{\circ }$. Therefore, the opposite angles at $K$ and $H$ of quadrilateral $FHEK$ are right angles and it is cyclic. Fig. 27