Let $B'=B_1H\cap A_1C_1$ and $C'=C_1H\cap A_1B_1$. Let $X$, $Y$ and $Z$ the midpoints of $AH$, $B'C'$ and $B_1C_1$ $\Longrightarrow$ it's well-know $X$, $Y$ and $Z$ are collinear and $XA_1^2=XY.XZ...(\star)$, so from $Y$ is the midpoint of $B'C'$ we get $AY$ is the A-symmedian of $\triangle A_1B_1C_1$ $\Longrightarrow$ $A$, $Y$ and $A_1$ are collinear. Since $B'C'B_1C_1$ is cyclic and $BC$ is tangent to $\odot (A_1B_1C_1)$ we get $\measuredangle QA_1B_1$ $=$ $\measuredangle B_1C_1A$ $=$ $\measuredangle C'B'A_1$ $\Longrightarrow$ $A_1Q\parallel B'C'$. Since $X$ is the center of $\odot (A_1C'B'HPQ)$ and $A_1Q\parallel B'C'$ we get $YX$ is bisector of $\measuredangle QXA_1$, but $XQ=XP$ combining both results we obtain $QXYP$ is cyclic. We consider the inversion $\mathbf{I}$ with center $X$ and radius $XA_1$ $\Longrightarrow$ by $(\star)$ we get $\mathbf{I} (Y)=Z$, since $QXYP$ is cyclic we get $\mathbf{I} (\odot (XYPQ))=\overline {ZPQ}$ $\Longrightarrow$ $Q$, $P$ and $Z$ are collinear hence $PQ$ bisects the line segment $B_{1}C_{1}$.

[asy][asy] unitsize(5cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(9); path carc(pair A, pair B, pair C, real d=0, bool dir=CW) { pair O=circumcenter(A,B,C); return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir); } pair A=dir(130), B=dir(210), C=dir(-30); path c=D(incircle(A,B,C),heavygreen+linetype("4 4")); pair A1=IP(c,B--C), B1=IP(c,C--A), C1=IP(c,A--B), H1=orthocenter(A1,B1,C1), E=foot(B1,C1,A1), F=foot(C1,A1,B1), P=foot(H1,A,A1), Q=foot(H1,B,C), M=(B1+C1)/2, N=(A1+H1)/2, X=(E+F)/2, Qp=2*N-Q; DPA(A--B--C--cycle^^E--F^^A--A1^^N--M^^P--H1--Q^^Qp--A1); D(A1--B1--C1--cycle,purple+linewidth(1)); D(CP(N,P),orange); D(carc(P,N,Q,47),red); D(carc(B1,E,C1,7),pink); D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("A_1",A1); D("B_1",B1,NE); D("C_1",C1,dir(C1)); D("E",E,dir(E)); D("F",F,dir(0)); D("H_1",H1,NW); D("P",P,NE); D("Q",Q); D("M",M,dir(90)); D("N",N,SE); D("X",X,SE); D("Q'",Qp,NE); [/asy][/asy] Let $E,F$ be the feet from $B_1,C_1$ to $C_1A_1,A_1B_1$. Let $M,X,N$ be the midpoints of $\overline{B_1C_1},\overline{EF},\overline{A_1H_1}$. Thus $N$ is the center of $(A_1EPH_1FQ)$, and denote this circle by $\omega$. Now $(B_1C_1EF),\omega$ are orthogonal, so $EF$ is the polar of $M$ wrt $\omega$: it follows that $X,M$ are inverses in $\omega$. Note that $EF$ is antiparallel to $B_1C_1$, so its midpoint $X$ lies on the symmedian $\overline{A_1A}$. We have $XN\perp EF\parallel BC$. Also, if $Q'$ is the reflection of $Q$ in $N$, we have $DQ'\parallel H_1Q\perp BC$. Thus $DQ'\parallel XN$, so by Reim's theorem, $PXNQ$ is cyclic. Now inverting in $\omega$, it follows that $P,M,Q$ are collinear, as required.

Let $ \triangle D_1E_1F_1 $ be the orthic triangle of $ \triangle A_1B_1C_1. $ Since $ H_1E_1, $ $ H_1F_1, $ $ H_1P, $ $ H_1Q $ is perpendicular to $ C_1A_1, $ $ A_1B_1, $ $ AA_1, $ $ BC, $ respectively, so $ H_1(E_1,F_1;P,Q) $ $ = $ $ -1 $ $ \Longrightarrow $ $ PE_1QF_1 $ is a harmonic quadrilateral, hence we conclude that $ PQ $ passes through the pole of $ E_1F_1 $ WRT $ \odot (A_1H_1). $ i.e. $ PQ $ passes through the midpoint of $ B_1C_1. $

Let $E,F$ be the feet of perpendicular from $B_1,C_1$ to $A_1C_1,A_1B_1$, respectively. Let $X=PA_1\cap EF$. $E,F,H_1,A_1,P,Q$ lie on a circle with diameter $A_1H_1$. There exists an inversion with center $A$ which maps $F\mapsto B_1,E\mapsto C_1$. Also, $AA_1$ is the $A_1$-symmedian in triangle $A_1B_1C_1$. Thus $X$ is the midpoint of $EF$. $\angle C_1A_1B=\angle C_1B_1A_1=\angle FEA_1$, so $EF\parallel BC\implies Q$ is the reflection of $A_1$ through the perpendicular bisector of $EF$. Thus $\frac{PE}{EQ}=\frac{PE}{FA_1}=\frac{PX}{XF}=\frac{PX}{XE}=\frac{PF}{EA_1}=\frac{PF}{FQ}$. Thus $EPFQ$ is harmonic quadrilateral $\implies PQ$ is the $P$-symmedian in triangle $PEF$. There exists a negative inversion with center $P$ which maps $B_1\mapsto E,C_1\mapsto F\implies PQ$ bisects $B_1C_1$.

Let $A_1D,B_1E,C_1F$ be altitudes of triangle $A_1B_1C_1$. Let $M$ be the midpoint of $B_1C_1$ and $P' \equiv QM \cap AA_1$. It is easy to see that $A_1,E,H,F$ lie on a circle with diameter $A_1H$. Because $\angle{HQA_1} = 90^{\circ}$, then $Q$ also lies on the circle with diameter $A_1H$. Notice also that we have $IA_1 \perp FE$ and $IA_1 \perp A_1Q$, so $A_1Q$ is parallel to $FE$, but because $A_1QFE$ are cyclic, then $A_1QFE$ must be an isosceles trapezoid, so $\triangle{QFE} \cong \triangle{AEF}$. It is well-known that $B_1FEC$ lie on a circle centered at $M$, so we get $\angle{MFE} = \angle{MEF} = \frac{180^{\circ} - \angle{EMF}}{2} = 90^{\circ} - \frac{\angle{EMF}}{2} = 90^{\circ} - \frac{2\angle{EC_1F}}{2} = 90^{\circ} - \angle{EC_1F} = \angle{C_1A_1F} \equiv \angle{EA_1F}$. From $\angle{MEF} = \angle{EA_1F}$, we get $ME$ and $MF$ are tangent to circle with diameter $A_1H$, and hence, $QM$ is the symmedian of $\triangle{QEF}$. Now, since $\triangle{A_1EF} \sim \triangle{A_1B_1C_1}$ and $\triangle{A_1EF} \cong \triangle{QFE}$, then $\triangle{QFE} \sim \triangle{A_1B_1C_1}$. Also, since $QA_1$ is parallel to $EF$ and $QH \perp QA_1$, we get $QH \perp EF$. Thus, we get \[\angle{HA_1A} = \angle{B_1AA_1} - \angle{B_1A_1H} = \angle{FQM} - \angle{FQH} = \angle{HQM}\]\[\angle{HA_1P'} = \angle{HQP'}\] Thus, $HQA_1P'$ is cylic and so $\angle{HP'A_1} = 90^{\circ}$ and this implies that $P' = P$. So, we are done.

Here is a solution with complex numbers that doesn't really have any computation. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair C,B,A,I,C1,B1,A1,H,P,Q,M,K,T; C=(1.22,0.51); B=(-5.78,0.51); A=(-4.986658647406665,3.391179561671051); I=(-4.207191860789772,1.7082660620805052); C1=(-5.36246214322365,2.02637316677807); B1=(-3.7026581108637755,2.795136454868686); A1=(-4.207191860789772,0.51); H=(-4.857928393297655,1.9149774974857456); P=(-4.605748003133087,1.9832017209893178); Q=(-4.857928393297655,0.51); M=(-4.532560127043713,2.410754810823378); K=(-4.989525958710855,2.1991041891544887); T=(-4.811326494546446,2.743091119677011); draw(A1--B1--C1--cycle); draw(A--A1); draw(P--H--Q,dashed); draw(Q--M,Dotted); draw(A--B--C--cycle); draw(incircle(A,B,C)); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$A_1$",A1,S); dot("$B_1$",B1,N); dot("$C_1$",C1,W); dot("$H$",H,dir(170)); dot("$P$",P,E); dot("$Q$",Q,S); dot("$M$",M,N); [/asy][/asy] Use complex numbers with $A_1=x$, $B_1=y$, $C_1=z$ and $(A_1B_1C_1)$ the unit circle, so that $H=x+y+z$, $A=\tfrac2{\frac1y+\frac1z}$. Let $\overline{A_1A}$ intersect $(A_1B_1C_1)$ at $T$, and let $M$ be the midpoint of $\overline{B_1C_1}$. The computation is relatively quick: \begin{align*} Q&=\frac12\left(2x+H-x^2\overline H\right) =\frac12\left(3x+y+z-x^2\left(\frac1x+\frac1y+\frac1z\right)\right) =x+\frac{y+z}2-\frac{x^2}A\\ P&=\frac{H+x+T(1-x\overline H)}2 =\frac{2x+y+z-x\frac{y+z}{yz}T}2 =x+\frac{y+z}2-\frac{xT}A.\\ \end{align*}Finally, \[\frac{Q-M}{P-M}=\frac{x-\frac{x^2}A}{x-\frac{xT}A}=\frac{1-\frac xA}{1-\frac TA}=\frac{A-x}{A-T}\in\mathbb R\]by $A$, $A_1$, $T$ collinear. Thus $P$, $Q$, $M$ collinear.