Problem 1869

Solution:

Since $\angle B A C$ and $\angle B D C$ subtend the same arc, we can let $\alpha=\angle B A G=\angle G A C=\angle C D G=\angle G D B$. Since $\angle B A G=\angle B D G$, then $G$ is a point on the circumcircle.
Let $x=\angle F H I$ and $y=\angle F I H$. Since $A H I D$ is cyclic $(\angle H A I=\angle I D H=\alpha)$, then $\angle I A D=x$ and $\angle H D A=y$. Since $A B C D$ is cyclic, we also have $\angle F B C=x$ and $\angle F C B=y$.
Since $F H G I$ is cyclic, then $\angle F G I=x$ and $\angle F G H=y$. By adding the angles of $\triangle A G D$, we get as a result: $x+y+\alpha=90^\circ$.
Extend $G F$, intersecting $A D$ at $J_{1}$, and the circumcircle of the pentagon at $E_{1}$. One consequence we get is that $G J_{1} \perp A D$ (because the highlighted angles of $\triangle A J_{1} G$, $\alpha+x+y$, already add up to $90^\circ$). Similarly, $D H \perp A G$ and $A I \perp D G$.

The equation now implies
$$\frac{J A}{J D}=\frac{F B}{F C} \cdot \frac{G I}{G H}=\frac{F H}{F I} \cdot \frac{G I}{G H}=\frac{F H / G H}{F I / G I}=\frac{F J_{1} / J_{1} D}{F J_{1} / J_{1} A}=\frac{J_{1} A}{J_{1} D}$$
This forces $J=J_{1}$ and so $E=E_{1}$. Therefore, $G, F$ and $E$ are collinear.