We state the following familiar lemma:
LEMMA. Given triangle $ABC$ and two points $X, Y$. Suppose $BX$ cuts $CY$ at $Z$ and $BY$ cuts $CX$ at $T$. Then if $AX$ and $AY$ are congruent in $\angle BAC$ then so are $AZ$ and $AT$.

We first reduce the problem to a simpler form through pairs of similar triangles. Let $D$ be the projection of $H$ on $BC$. It is easy to see that
$$\triangle MFE^{UK} \sim \triangle OBC^{UD}$$
so we think of constructing triangle $TBC$ similar to triangle $RFE$ where $RE$ and $RF$ are tangent to $(PBF)$ and $(QCE)$.

Let $X, Y$ be the intersection of $AP$ and $AQ$ with $EF$. Firstly
$$\angle XPF = 90^\circ - \angle OAX = \angle PBF$$
infer $X$ is on $(BPF)$ or
$$\angle RFE = \angle BXA \text{ and } \angle REF = \angle YCA.$$

Therefore, if $\triangle TBC \sim \triangle RFE$ then $BT$, $BX$ is isogonal in $\triangle ABC$ and $CY$, $CT$ is isogonal in $\triangle ACB$. In short, if $S$ is the intersection of $BX$ and $CY$, then $S$, $T$ is isogonal conjugate in triangle $ABC$.

Let $L$ be the intersection of $BY$ with $CX$ and $J$ the isogonal conjugate of $L$ in triangle $ABC$. We will prove that $O$ and $D$ lie on $JT$. Indeed,
$$\begin{aligned} B(JT, OC) &= B(LS, HA) = (YX, EF) = C(YX, EF) \\ &= C(XY, FE) = C(LS, HA) = C(JT, OB) \end{aligned}$$
so $J$, $O$, $T$ are collinear.

Next, since $AX$ and $AY$ are isogonal in $\angle BAC$, according to the lemma, $AS$, $AL$ isogonal in $\angle BAC$ leads to $A$, $L$, $T$ and $A$, $J$, $S$ are collinear. Furthermore, in the complete quadrilateral $BLSC.XY$, if $XY$ intersects $BC$ at $D'$ then
$$L(CB, SD') = -1 = L(CB, DD'),$$
in other words, $SL$ goes through $D$. Finally, to prove that $JT$, $SL$ and $BC$ are concurrent, we use Desargues' theorem for the pair of triangles $BLT$ and $CSJ$. Since $BT$, $BX$ and $CJ$, $CX$ are isogonal pairs in triangle $ABC$, the intersection $Z$ of $BT$ and $CJ$ is the isogonal conjugate of $X$ in the triangle $ABC$, thus $Z$ is on $AY$. Therefore, the intersection of $(BT, CJ)$, $(BL, CS)$ and $(TL, JS)$ are collinear or $JT$ passes through the point $D$.

Therefore, $T$ belongs to $OD$ so $R$ belongs to $MK$. This finishes the proof. $\Box$