Mathematica competitions in Croatia

Let us denote $\angle BAC=\alpha$, $\angle CBA=\beta$, $\angle ACB=\gamma$, and let $J$ be the intersection of the segments $\overline{DE}$ and $\overline{CB}$. Inscribed angles $\angle BED$, $\angle BCD$ and $\angle BAD$ over the chord $BD$ are equal, so $\angle BED=\angle BCD=\angle BAD=\frac{\alpha }{2}$. Analogously, $\angle DEC=\angle DBC=\angle DAC=\frac{\alpha }{2}$, $\angle ECA=\angle EBA=\frac{\beta }{2}$ and $\angle CDE=\angle CBE=\frac{\beta }{2}$. Since $\angle DFC=\angle DEC+\angle ECA=\frac{\alpha }{2}+\frac{\beta }{2}$ and $\angle CJE=\angle CDE+\angle BCD=\frac{\beta }{2}+\frac{\alpha }{2}$, the triangle $CFJ$ is isosceles. On the other hand, lines $FH$ and $EG$ are parallel, thus $\angle DHF=\angle DGE=180^{\circ }-\angle ECD=180^{\circ }-(\angle ECA+\angle ACB+\angle BCD)=180^{\circ }-(\frac{\beta }{2}+\gamma +\frac{\alpha }{2})=\frac{\alpha }{2}+\frac{\beta }{2}=$

$<DFC$ and, by the converse of the tangent chord theorem, we conclude that the line $CA$ touches the circumcircle of the triangle $DFH$ at point $F$. Since line $CI$ is the angle bisector of $<FCJ$, we have $CI\perp DE$. Since $<BED$ = $<DEC$, line $DE$ is the bisector of the segment $\overline{CI}$ and $<IFD$ = $<DFC$ = $\frac{\alpha }{2}+\frac{\beta }{2}$. This implies that $<HFD$ = $<DFC$ = $<CJE$ = $<BJD$. We conclude that the lines $FH$ and $BC$ are parallel. From $<FHD$ = $<GED$ = $\frac{\alpha +\beta }{2}$ = $<HFD$ we conclude that the triangle $DFH$ is isosceles. Since $D$ is the midpoint of the arc $\text{overarc}BC$ and triangles $DCB$ and $DFH$ are isosceles with $BC$ parallel to $FH$, the line $BH$ is symmetric to the line $CF$ with respect to the bisector of the segment $\overline{FH}$. Hence, the line $BH$ also touches the circumcircle of the triangle $DFH$, at point $H$.