Pythagoras/quadratic equations are illegal (but power of points).
$\triangle ABC$ is equilateral, $ACED$ is a rectangle, $G$ is the center of $\triangle ABC$. It's not hard to see that $H$ is midpoint of $AC$ and $F$ is midpoint of $ED$.
I also know that $AG_1CB$ is harmonic and I thought (incorrectly) that because $FA=FC$ and $AG_1CB$ is harmonic, then $AC$ should be polar of $F$ but althought the later result is true, this logic is false. Is there a nice way to see that $AF$ is tangent to the circle without calculating stuff?
The complete quadrangle $DEXY$ has the property that one pair of opposite sides intersect at $B$, a second pair intersect at $G_1$, and the third pair meet $BG_1$ at $F, H$. Therefore, $FG_1HB$ is a harmonic range—indeed, the existence of such a quadrangle is often taken as the definition of a harmonic range.
This immediately proves the statement in your question title, but of course it ignores most of your diagram—which doesn’t make it wrong, but does make me wonder whether you’re working with a different definition.