We claim that the point of concurrency is $H$, the orthocenter of triangle $ABC$. By symmetry, it suffices to show that the perpendicular from $D$ to $AA_2$ passes through $H$.

Let $A_3$ be the projection of $D$ onto $AA_2$. Because $\angle AA_1D = \angle AA_3D = 90^\circ$ and $\angle BC_1C = \angle BB_1C = 90^\circ$, quadrilaterals $AA_3A_1D$ and $BC_1B_1C$ are cyclic. Notice now that points $D$, $E$, $F$, $A_1$, $B_1$, and $C_1$ lie on the nine-point circle of triangle $ABC$. Further, by Power of a point on cyclic quadrilaterals $AA_3A_1D$, $A_1DC_1B_1$, and $BCB_1C_1$, we obtain
$$A_2A_3 \cdot A_2A = A_2A_1 \cdot A_2D = A_2B_1 \cdot A_2C_1 = A_2B \cdot A_2C.$$
By the converse of Power of a point, it follows that $A_3$ lies on the circumcircle $\omega$ of $ABC$.

Extend segment $AA_1$ through $A_1$ to meet $\omega$ at $H_2$. Then
$$\angle A_1 BH_2 = \angle CBH_2 = \angle CAH_2 = \angle CAA_1 = 90^\circ - \angle ACB = \angle B_1 BC = \angle HBA_1.$$
That is, in triangle $HBH_2$, segment $BA_1$ bisects $\angle HBH_2$ and is the altitude from $B$ to side $HH_2$. Hence, $HBH_2$ is isosceles and $HA_1 = A_1H_2$.

Reflect $H$ through $D$ to obtain $H_3$. Then $DA_1$ is the midline of right triangle $HH_2H_3$. Let $M$ be the midpoint of $H_2H_3$. Then $DM$ is a midline of right triangle $HH_2H_3$. In particular, $DM \perp BC$ and so $DM$ passes through the circumcenter $O$ of triangle $ABC$. Because $MO$ is the perpendicular bisector of segment $H_2H_3$ and $H_3$ lies on $\omega$, $H_2$ also lies on $\omega$. Because $\angle AH_2H_3 = 90^\circ$, $AH_3$ is a diameter of $\omega$, which implies that $\angle AA_3H_3 = 90^\circ$ and hence $H$ lies on $DA_3$, as needed.