Croatian Mathematical Competitions

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 $\text{BC}$ 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 $H$.