Let $M_1$, $M_2$, and $M_3$ be the midpoints of $CE$, $AE$, $AC$ and $N_1$, $N_2$, and $N_3$ be the midpoints of $DF$, $BF$, and $BD$. Also, let $H$ be the orthocenter of $XYZ$. Note that we can use parallel sides to see that $X$, $Z$, and $M_3$ are collinear. Thus we have $$\text{Pow}(M_3,(XYZ)) = M_3Z \cdot M_3X = \frac 14 AB \cdot DE$$ by midlines. Applying this argument cyclically, and noting the condition $AB \cdot DE = BC \cdot EF = CD \cdot FA$, $M_1$, $M_2$, $M_3$, $N_1$, $N_2$, $N_3$ all lie on a circle concentric with $(XYZ)$.
Next, realize that basic orthocenter properties imply that the circumcenter $O_1$ of $(ACE)$ is the orthocenter of $\triangle M_1M_2M_3$, and likewise the circumcenter $O_2$ of $(BDF)$ is the orthocenter of $\triangle N_1N_2N_3$.
The rest is just complex numbers; toss on the complex plane so that the circumcenter of $\triangle XYZ$ is the origin. Then we have $$o_1 = m_1+m_2+m_3 = (c+e)/2+(a+e)/2+(a+c)/2=a+c+e$$ $$o_2 = n_1+n_2+n_3 = (b+d)/2+(d+f)/2+(b+f)/2=b+d+f$$ $$h = x+y+z = (a+d)/2+(b+e)/2+(c+f)/2=(a+b+c+d+e+f)/2.$$
Note that from the above we have $h=\frac{o_1+o_2}{2}$, so $H$ is the midpoint of segment $O_1O_2$. In particular, $H$, $O_1$, and $O_2$ are collinear, as required.
~ Leo.Euler