Wlog suppose $\angle B > \angle C$. (In the case that they are equal, we are done by symmetry). Take a homothety at the centroid with ratio $-2$. The midpoint of the $A-$altitude goes to the reflection of the foot of the $A-$altitude in the midpoint of $BC$ (considering the anticomplementary triangle), say $D'$. Define $E',F'$ similarly. The parallels to $OA, OB, OC$ are merely the perpendiculars to the sides of the tangential triangle $UVW$. Hence we only need to show that $D'E'F'$ and $UVW$ are orthologic. So we need to show that the perpendiculars from $U, V, W$ to the sides of $D'E'F'$ are concurrent. We claim that the required concurrency point is the Nagel point of $UVW$. Suppose $L$ is the midpoint of $VW$. Then since the $OL$ is parallel to the $U-$Nagel cevian, we need to show that $OL \perp E'F'$. Suppose the pole of $E'F'$ wrt $(ABC)$ is $K$, then it suffices to show that $O, L, K$ are collinear. Since $LA$ is tangent to $ABC$, and since the other tangent from $L$ meets $(ABC)$ on the $U-$Nagel cevian, which also passes through the antipode $N$ of $A$ in $(ABC)$, it suffices to show that if $UN$ meets $(ABC)$ again at $X$, then $AX \parallel E'F'$. Suppose $A'$ is the reflection of $A$ in the perpendicular bisector of $BC$. Then we have $\frac{BF'}{BA'} = \frac{AF}{AC} = \frac{AE}{AB} = \frac{CE'}{CA'}$ and $\angle A'CE = \angle A'BF$, which means $A'CE$ and $A'BF$ are similar, and thus $A'$ is the center of spiral similarity sending $BF'$ to $CE'$, and thus $A'AE'F'$ is cyclic. By Reim's theorem, the line through $C$ and parallel to $E'F'$ meets $(ABC)$ again on $A'F'$, at say $R$. Note that if the Miquel point of $EFBC$ is $I$, then $UI$ meets $ABC$ on $AD$, hence by reflection under the perpendicular bisector of $BC$, $IX \parallel BC$. We show that the mentioned parallel line would be isogonal to $CI$ in $\angle ACB$, for this would imply that $\angle AXI = \angle ACI = \angle RCB = \angle(RC, IX) = \angle(E'F', IX)$, whence we would be done. By reflection under the perpendicular bisector of $AB$, it suffices to show that $IF$ passes through the reflection of $A'$ in the perpendicular bisector of $AB$, which is equivalent to showing that $\angle BIF = B - C$. However this is obvious as $I$ is the center of spiral similarity sending $FE$ to $BC$, whose angle is $B - C$, and we are done.

A less high-powered solution (sorry for the double post): Note that the line through the foot of $A$ onto $BC$ which is parallel to $OA$ passes through the orthocenter of the orthic triangle, so the required concurrence point is in fact the midpoint of $O$ and the said orthocenter.