Let $O_1, O_2, O_3$ and $O_4$ be the circumcentres of the triangles $ADE$, $ADC$, $BFD$ and $BCD$. The line $O_1O_2$ bisects the segment $AD$ and the two are perpendicular. Similarly, $O_3O_4$ bisects the segment $DB$ and these two are perpendicular as well.

Denote the angles of the triangle by $\alpha$, $\beta$ and $\gamma$ and let $T_1, T_2, T_3, T_4$ and $T_5$ be the midpoints of the segments $AD$, $CF$, $BD$, $CE$ and $CD$.
We will be using directed angles as this will shorten the calculation. The quadrilateral $ADFC$ is cyclic, so $\angle DFB = \angle DFC = \angle DAC = \alpha$. Since $O_3$ is the circumcentre of the triangle $DBF$ and $DBF$ is an acute triangle, we have $\angle DO_3T_3 = \angle DFB = \alpha$. Since $O_2$ is the circumcentre of the triangle $ADC$, we have $\angle DO_2T_5 = \angle DAC = \alpha$ (we used the fact that $\angle DAC$ is an acute angle).
The points $O_2, T_5$ and $O_4$ are collinear, so $\angle DO_2O_4 = \angle DO_2T_5 = \alpha = \angle DO_3T_3 = \angle DO_3O_4$ and $\angle DO_2O_4 = \angle DO_3O_4$. Hence, the points $O_2, O_3, O_4$ and $D$ are concyclic.
A similar argument (but for the angle $\beta$) shows that $O_4, D, O_1$ and $O_2$ are concyclic. Thus, $O_1$ and $O_3$ lie on the circuncircle of the triangle $O_2O_4D$. We know that $\angle DO_2O_4 = \alpha$ and $\angle O_2O_4D = \beta$. So $\angle O_4DO_2 = \gamma$. On the other hand, the quadrilateral $CT_4OT_2$ is cyclic, so $\angle T_4OT_2 = \angle T_4CT_2 = \gamma$. Since $T_2, O, O_2$ are collinear and $T_4, O, O_4$ are collinear, we get $\angle O_4OO_2 = \angle T_4OT_2 = \gamma = \angle O_4DO_2$ and $O$ lies on the circuncircle of the triangle $O_2DO_4$. Hence, the points $O_1, O_2, O_3, O_4, O$ and $D$ are concyclic.

We have
$$\begin{aligned} \angle OO_4O_3 &= \angle T_4O_4E + \angle EO_4D + \angle DO_4T_3 \\ &= \angle CBE + \angle EBD + \angle EBD + \angle DEB \\ &= \angle CBD + \angle EDB \\ &= \beta + \gamma \end{aligned}$$
and $\angle DOO_4 = \angle DO_3O_4 = \alpha$. So $OD$ is parallel to $O_3O_4$, which is perpendicular to $AB$.