Given a square $ABCD$, as shown. We choose a point $E$ on the side $CD$. Draw a line through $D$ which is parallel to $BE$ and let $X$ be the intersection of this line with $AB$. Draw a line through $E$ which is parallel to $BC$ and let $H$ be the intersection of this line with the diagonal $AC$. Let $F$ be the intersection of $EH$ and $DX$ and, draw a line through $F$ which is parallel to $CD$ and let $G$ be the intersection of this line with $AD$. Now draw a line through $G$ which is parallel to $DX$. Prove that this line passes through $H$ $\textit{iff}$ $$\frac{EC}{CB} = \sqrt{3}-1.$$
Idea: Since we are to prove collinearity, I think I will need to use Menelaus theorem. But I seem not to be able to find a suitable one in the diagram.
All that needs to be shown is $GH\parallel DX$.
Set up a coordinate system with $B=(0,0)$, $C=(1,0)$, $A=(0,1)$ and $E=(1,k)$. $DX$ has equation $y=kx+(1-k)$, while $EH$ has equation $y=k$. Intersecting $DX$ and $EH$ gives the coordinates of $F$ and thus $G$: $$F=(2-1/k,k)\qquad G=(2-1/k,1)$$ $H$ has coordinates $(1-k,k)$. Now consider the slope of $GH$, equate it to that of $DX$ ($k$) and solve: $$\frac{1-k}{2-1/k-(1-k)}=k$$ $$1-k=k-1+k^2$$ $$k^2+2k-2=0$$ $$k=-1\pm\sqrt3$$ Since $0\le k\le1$, we must have $k=\sqrt3-1$, which completes the proof.