We apply barycentric coordinates with respect to $\triangle ABC$. Let $D=(p,q,r)$ [with $p+q+r=1$]; then \[E=(p:0:r)=\left(\frac p{p+r},0,\frac r{p+r}\right), F=(p:q:0)=\left(\frac p{p+q},\frac q{p+q},0\right);\]Now we have \[X=B+E-F=(p(q-r),p(p+r),r(p+q))\]and thus by symmetry \[Y=C+F-E=(p(r-q):q(p+r):p(p+q)).\]It remains to find the ratio of $x,y$ coordinates of $T=\overline{EY}\cap\overline{FX}$; let $T=(f:g:h)$, so that \[ \begin{vmatrix}f&g&h\\ p&0&r\\p(r-q)&q(p+r)&p(p+q)\end{vmatrix} =0= \begin{vmatrix}f&g&h\\ p&q&0\\p(q-r)&p(p+r)&r(p+q)\end{vmatrix}, \]which gives the system (after cancelling factors of $p+r$ and $p+q$ for the equations respectively) \begin{align*} qrf+p(p+q-r)g-pqh&=0;\\ qrf-prg+p(p+r-q)h&=0. \end{align*}Subtracting these and rearranging gives \[\frac g h=\frac{p+r}{p+q},\]which is equal to \[\frac{b^2/c^2}{q/r}\]iff $AE/AF=c/b$: $AE=br/(p+r),AF=cq/(p+q)$. But $BCEF$ cyclic iff $AE/AF=c/b$, so we're done! Note: Cancellations of $p,p+q,p+r$ permissible because they're all strictly positive by the problem conditions.