In Triangle $ABC$, $D$ is on $AB$, $E$ is on $AC$, $CD$ meets $BE$ at $J$. $F$ is a moving point between $BC$. $FG \parallel CD$, $FH \parallel BE$. The circle centered at $G$, with radius $GB$, has a common chord with the circumcircle of $\triangle AGH$ (the light line on the grapch). Prove that the common chord passes a fixed point.
My thoughts: try two special points of $G$ when $G$ is the midpoint of $AB$, and when $G$ is at somewhere else. Then the intersection of the common chord should be the fixed point we are looking for. But didn't go very far with that
I would start by looking at the extreme cases. If $F=B$ then $AGH=ABE$ so the radical axis would be the tangent to $(ABE)$ through $B$. If the fixed point exists it must be on that tangent. Now, when $F=C$ it is a little more complicated. One gets the radical axis of $(ACD)$ and the circle of radius $DB$ centered at $D$. I don't know how to construct that nicely but at least we know its direction: it's perpendicular to $DO$ where $O$ is the center of $(ACD)$.
However, another idea worked for me: Prove that $(AGH)$ passes through another fixed point besides $A$ (it should be clear which point by considering extreme cases as above). And now you can restate the problem:
Suppose we have two circles $\Gamma_1$ and $\Gamma_2$ intersecting at $A$ and $A'$. Let $B$ be a point on $\Gamma_1$ and $\Gamma$ any circle passing through $A$ and $A'$. Let $G$ the intersection of the line $AB$ with $\Gamma$. Prove that the radical axis of $\Gamma$ and the circle of radius $GB$ centered at $G$ passes through a fixed point.
(Edit: notice $\Gamma_2$ is of no use, I just left that for clarity, to see how it relates to the previous problem. So it's even simpler, $A,B,A'$ are fixed points on $\Gamma_1$ and what changes is $\Gamma$.)
That happens to be the case, but still some work remains to be done.