, 2023

Solution. We first prove that D', H, X are collinear: since $\angle GXB = \angle ACB = \angle GFB$, we get that G, F, B, X are concyclic.
Combining this with $AG \cdot AX = AF \cdot AB = AH \cdot AD$, we obtain that G, X, D, H are concyclic. Note that $\triangle DFH \sim \triangle DAE$ and $\triangle DFG \sim \triangle DD'E$, hence $\triangle DGH \sim \triangle DAD'$, so $\angle DXA = \angle DHG = \angle DD'A$, that is, A, X, D, D' are concyclic. This tells us that $\angle DXH = \angle DGH = \angle DAD' = \angle DXD'$, that is, D', H, X are collinear.
Since AH, BC are the two angle bisectors of $\angle GDD'$, we know that $(P,Y;G,D') = (AH,BC;DG,DD') = -1$, where P is the intersection point of AH and GD'. Therefore
$$A(X, Z; Y, H) = (G, D'; Y, P) = -1 = (D', G; Y, P) = H(X, Z; Y, A),$$
that is, X, Y, Z are collinear. $\square$