$$\frac{HQ}{PA} = \frac{HS}{AS} \quad \text{and} \quad \frac{RH}{PA} = \frac{HD}{AD}.$$
It suffices to show that
$$\frac{HS}{AS} = \frac{HD}{AD}, \qquad (29)$$
which we remark is equivalent to showing that $(A, H)$ and $(S, D)$ are harmonic conjugates. Applying Ceva's theorem to triangle $AHC$ and cevians $AD$, $BE$, and $CF$ gives
$$\frac{AE}{EC} \cdot \frac{CF}{FH} \cdot \frac{HD}{DA} = 1.$$
Applying Menelaus's theorem to triangle $ACH$ and line $ESF$ yields
$$\frac{AE}{EC} \cdot \frac{CF}{FH} \cdot \frac{HS}{SA} = 1.$$
Equating the left-hand sides of the last two equations gives (29), completing the proof.

Consider the diagram shown above. We present another proof of (29). Let $\angle ABC = B$ and $\angle ACB = C$. In right triangles $BHD$ and $ABD$, we have $HD/BD = \tan \angle HBD = \cot C$ and $AD/BD = \tan B$. Dividing the last two equations gives
$$\frac{HD}{AD} = \frac{\cot C}{\tan B} = \cot B \cot C. \qquad (30)$$
Applying the Law of Sines in triangles EHS and AFS gives
$$\frac{HS}{HE} = \frac{\sin \angle SEH}{\sin \angle HSE} \quad \text{and} \quad \frac{AS}{AF} = \frac{\sin \angle AFE}{\sin \angle ASF}.$$
Dividing the last two equations yields
$$\frac{HS}{AS} = \frac{HE}{AF} \cdot \frac{\sin \angle SEH}{\sin \angle AFE}. \qquad (31)$$
Because $\angle BEC = \angle CFB = 90^\circ$, $BCEF$ is cyclic, from which it follows that $\angle SEH = \angle FEB = \angle FCB = 90^\circ - B$ and $\angle AFE = \angle ECB = C$. Substituting the last two equations into (31) yields
$$\frac{HS}{AS} = \frac{HE}{AF} \cdot \frac{\cos B}{\sin C}. \qquad (32)$$
It is clear that right triangles CHE and CAF are similar to each other. Hence
$$\frac{HE}{AF} = \frac{CE}{CF} = \frac{BC \cos C}{BC} \sin B = \frac{\cos C}{\sin B}.$$
Substituting the last equation into (32) gives
$$\frac{HS}{AS} = \frac{HE}{AF} \cdot \frac{\cos B}{\sin C} = \frac{\cos C}{\sin B} \cdot \frac{\cos B}{\sin C} = \cot B \cot C.$$
Combining the last equation with (30) gives (29), and we are done.

Consider the diagram shown above. Because $\angle HFB + \angle HDB = 180^\circ$, $BDHF$ is cyclic, from which it follows that $\angle HDF = \angle HBF$. Likewise, $\angle EDH = \angle ECH$. Because $\angle BFC = \angle BEC = 90^\circ$, $BCEF$ is cyclic, from which it follows that $\angle EBF = \angle ECF$. Therefore, we have $\angle HDF = \angle HBF = \angle EBF = \angle ECF = \angle ECH = \angle EDH$; that is, $DH$ bisects $\angle FDE$. Likewise, $FH$ bisects $\angle EFD$ and $EH$ bisects $\angle DEF$. Hence $H$ is the incenter of triangle $DEF$. Let $\omega$ denote the incircle of triangle $DEF$.
Because $\angle AE \perp EH$ and $EH$ is the interior bisector of $\angle DEF$, $AE$ is the exterior bisector of $\angle DEF$. Likewise, $AF$ is the exterior bisector of $\angle EFD$. Consequently, $A$ is the excenter (opposite $D$) of triangle $DEF$. Let $\Omega$ denote the excircle of triangle $DEF$ centered at $A$.
Consider the dilation centered at $D$ sending $\omega$ to $\Omega$. Because $HQ \parallel AP$, line $QH$ is sent to line $AP$. Let $R_1$ be the point on $\omega$ diametrically opposite $Q$ (so $QH = HR$ and $Q, H, R_1$ are collinear). Then, we see that $R_1$ is sent to $P$ by this dilation. In particular, $D, R_1, P$ are collinear. Hence, $R_1$ is the intersection of lines $DP$ and $QH$, so $R = R_1$. We therefore have that $RH = R_1H = QH$.