If N is the midpoint of AH, where H is orthocentre, then $A_0N$ is angle bisector along with AN=$NB_1$ leads to conclusion

It is well known that $C_1HB_1A$ is cyclic with $A_0C_1$ and $A_0B_1$ as tangents. [asy][asy] import olympiad; import geometry; size(300); pair A,B,C,B1,C1,H,A0,I,B2,C2; A=(5,6); B=(0,0); C=(7,0); B1=foot(B,A,C); C1=foot(C,A,B); H=orthocenter(A,B,C); I=(A+H)/2; A0=(B+C)/2; B2=(-3,6); C2=(15,6); pair P=intersectionpoint(line(B2,C2),line(A0,B1)); pair Q=intersectionpoint(line(B2,C2),line(A0,C1)); draw(P--I); draw(Q--I); draw(B2--C2); draw(A--B--C--cycle); draw(B--B1); draw(C--C1); draw(A--H); draw(I--B1); draw(I--C1); draw(A0--Q); draw(A0--P); draw(circumcircle(A,I,C1),red+dotted); draw(circumcircle(A,I,B1),blue+dotted); dot("$A$",A,N); dot("$B$",B,SW); dot("$H$",H,S); dot("$C$",C,SE); dot("$B_1$",B1,NE); dot("$C_1$",C1,NW); dot("$I$",I,E); dot("$A_0$",A0,S); dot("$P$",P,N); dot("$Q$",Q,N); [/asy][/asy] Let $\angle ABC=B,\angle BAC=A,\angle ACB=C$. Let $l$ be the line parallel to $BC$ through $A$. Let $I$ be a point on $AH$ such that $PI$ is the angle bisector of $\angle QPA_0$. We know $A_0B_1=A_0C=A_0B=A_0C_1$. So, $\angle B_1BA_0=\angle BB_1A_0=90^\circ-C$. So, $$\angle APA_0=\angle CA_0B_1=180^\circ-2C \implies \angle API=\angle IPA_0=90^\circ-C=\angle A_0B_1B\implies BB_1\parallel PI.$$Also, $\angle IAB_1=90^\circ-C=\angle IPB_1$. So, $IAB_1P$ is cyclic and hence $IA=IB_1$. So, $I$ is the midpoint of $AH$. Let $Q'$ be a point on $l$ such that $IQ'\parallel CC_1$. So, $$\angle AQ'I=\angle A_0CC_1=\angle A_0C_1C=\angle C_1AH=\angle AC_1I.$$Hence, $C_1IAQ'$ is cyclic. So, $\angle C_1Q'I=\angle C_1AI=\angle AQ'I$. So, $Q'I$ is angle bisector of $\angle AQ'C_1$ and also $A_0,C_1,Q'$ are collinear and hence $Q'\equiv Q$. Hence, we are done.$\blacksquare$.

@above There is a much quicker solution once you remember the lemma regarding tangents to $(AC_{1}HB_{1})$. By that lemma both $QC_{1}$ and $AQ$ are tangents to that circle. So $QC_{1}=QA$, let $I$ be the center of $(AC_{1}HB_{1})$. By quick angle chasing, $QI$ is the angle bisector of $\angle C_1QA$. Analogously $PI$ is the angle bisector of $\angle B_{1}PA$. $I$ is the incenter of $\triangle PQA_{0}$, and $I \in AH$.

@abovex2 Well, we know $(C_1HB_1A)=\omega$ is cyclic and also the center is on $AH $. $A_0C_1$ and $A_0B_1$ are tangents to $\omega $ and notice that $AH\perp BC \implies AH\perp PQ $ yielding $PQ $ is also tangent to $\omega $ this implies $\omega $ is the incenter of $A_0PQ $ and the result follows.

homothety solution: Let $I$ be the incenter of $\triangle PA_0Q$. Note that points $B,C,B_1,C_1$ lie on a circle centered at $A_0$. This gives angles of $\triangle A_0QP$ are $180^\circ - 2A, 180^\circ - 2B, 180^\circ - 2C$. Then we easily have $\overline{QI} \perp \overline{AC_1}$ and $\overline{PI} \perp \overline{AB_1}$. Let $X = \overline{PI} \cap \overline{AB_1}$ and $Y = \overline{QI} \cap \overline{AC_1}$. [asy][asy] size(200); pair A = dir(100),B=dir(-150),C=dir(-30),A0=0.5*(B+C),B1 = foot(B,A,C), C1 = foot(C,A,B), I = circumcenter(A,B1,C1),Q=extension(A0,C1,I,foot(I,A,B)),P=extension(A0,B1,I,foot(I,A,C)),X=foot(I,A,C),Y=foot(I,A,B), Ip = extension(B,B1,C,C1); fill(A--Q--Y--A--cycle,green+grey+grey); fill(C1--Q--Y--C1--cycle,green+grey); fill(P--A--X--P--cycle,purple+grey+grey); fill(B1--P--X--B1--cycle,purple+grey); fill(I--Y--X--I--cycle,red+grey+grey); fill(Ip--B1--C1--Ip--cycle,red+grey+grey); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A_0$",A0,dir(A0)); dot("$B_1$",B1); dot("$C_1$",C1,dir(C1)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$I$",I,dir(-60)); dot("$X$",X,dir(-105)); dot("$Y$",Y,dir(-73)); dot("$I^*$",Ip,dir(-90)); draw(A--B--C--A,fuchsia); draw(C--C1^^B--B1,fuchsia); draw(Q--A0--P--Q,royalblue); draw(Q--I--P^^A--Ip,orange); draw(X--Y^^B1--C1,orange); [/asy][/asy] Then by congruent triangles we have $X,Y$ are midpoints of segments $AB_1,AC_1$, respectively. Now consider the homothety at $A$ sending $X \to B_1$ and $Y \to C_1$ (suppose it sends $I$ to $I^*$. Then $I^*$ is a point such that $\overline{I^*C_1} \perp \overline{AB}$ and $\overline{I^* B_1} \perp \overline{AC}$, so $I^*$ is the orthocenter of $\triangle ABC$, meaning lines $AI$ passes through orthocenter of $\triangle ABC$, and we are done. $\blacksquare$

Solution using angle chasing We first perform some elementary angle chasing. Since $\triangle BC_1C$ is right angled, $\angle BA_0C_1 =180- 2B = \angle C_1QA$. But $\angle QC_1A =\angle BC_1A_0 =B$. So $\triangle QAC_1$ is isosceles. Similarly, $\triangle APB_1$ is also isosceles. Now $\angle C_1A_0B_1= 180-(180-2B+180-2C) =180-2A$. Also since $AC_1B_1 =C$ and $\angle AB_1C_1 = B$. So $\triangle A_0B_1C_1$ is also isosceles. Now consider the circle $\odot AC_1HB_1$. Using tangent chord theorem , multiple times on this circle, we get that this circle is tangent to all the sides of the desired triangle($\triangle PA_0Q$). So the incentre of this triangle is the centre of the circle, which is the midpoint of $AH$, and we're done $\blacksquare$.

A Very Quick Solution By the three tangents lemma we know $\odot(BC_1B_1C)$ with centre $A_0$. Now by symmetry (or just congruence) $QA = QB_1$ and similarly $PA = PC_1$ and $A_0B_1=A_0C_1$ which implies $\odot(AB_1C_1)$ is the incircle of $\triangle PA_0Q$ which is the midpoint of $AH \ \blacksquare$

Let $H$ be the orthocenter of $ABC$. The Three Tangents Lemma implies $A_0B_1, A_0C_1, PQ$ are each tangent to $(AB_1HC_1)$, i.e. $AB_1C_1$ is the contact triangle of $A_0PQ$. To finish, we notice that Thales' yields the circumcenter of $AB_1C_1$ as the midpoint of $AH$. $\blacksquare$

parmenides51 wrote: Let $BB_1$ and $CC_1$ be the altitudes of acute-angled triangle $ABC$, and $A_0$ is the midpoint of $BC$. Lines $A_0B_1$ and $A_0C_1$ meet the line passing through $A$ and parallel to $BC$ in points $P$ and $Q$. Prove that the incenter of triangle $PA_0Q$ lies on the altitude of triangle $ABC$. Trivial!,by 3-TL $A_0B_1=A_0C_1,QC_1=QA,PB_1=PA$ So $(\triangle AB_1C_1)$,is the incircle and we are done. Bruuh