Junior Balkan Mathematical Olympiad Shortlist

Solution:
We can see easily that points $C$, $D$, $H$, $F$ lie on a circle of diameter $[CH]$.
Take $\{F, G'\} = \odot(CHF) \cap EF$. We have $\widehat{EFH} = \widehat{BAD} = \widehat{BCE} = \widehat{DFH}$ since the quadrilaterals $AEDC$, $AEHF$, $CDHF$ are cyclic. Hence $FB$ is the bisector of $\widehat{EFD}$, so $H$ is the midpoint of the arc $DG'$. It follows that $DG' \perp CH$ since $[CH]$ is a diameter. Therefore $DG' \parallel AB$ and $G \equiv G'$. Finally $G$ lies on the circle $\odot(CFH)$, so $\widehat{HGC} = 90^\circ$.

Solution:
The quadrilateral $AEHF$ is cyclic since $\widehat{AEH} = \widehat{AFH} = 90^\circ$, so $\widehat{EAD} \equiv \widehat{GFH}$.
But $AB \parallel GD$, hence $\widehat{EAD} \equiv \widehat{GDH}$. Therefore $\widehat{GFH} \equiv \widehat{GDH} \Rightarrow DFGH$ is cyclic.
Because the quadrilateral $CDHF$ is cyclic since $\widehat{CDH} = \widehat{CFH} = 90^\circ$, we conclude that the quadrilateral $CFGH$ is cyclic, which gives that $\widehat{CGH} = \widehat{CFH} = 90^\circ$.