Let $CH\cap AB = F$ and $BH\cap AC = E.$ Also, let $M$ be the midpoint of $AB$ and let $N$ be the midpoint of $AC.$ Note that $(MFEND)$ is the nine-point circle of $\triangle ABC,$ which we will denote with $\Gamma.$ Claim $1:$ $P,H,N$ are collinear. Proof: Let $B'$ be the antipode of $B$. It is well-known that $H,N,B'$ are collinear, so if $HN$ intersects $(ABC)$ for a second time at point $P_1$, then $\angle BP_1H=90^{\circ}.$ However, we know that $\angle BPH=90^{\circ},$ so $P\equiv P_1$ and we are done. Similarly, it follows that $Q,H,M$ are collinear. Now consider inversion $\Psi$ centered at $H$ with radius $R=-\sqrt{HA\cdot HD.}$ Note that under this inversion, we have that $\Psi(D)=A, \Psi(E)=B$ and $\Psi(F)=C$, so $\Psi(\Gamma)=(ABC).$ Hence, $\Psi(P)=N$ and $\Psi(Q)=M$. Let $\Psi(X)=Y.$ Since $X=OH\cap BC$, we get that $Y=\Psi(OH)\cap\Psi(BC)=OH\cap (HEF).$ Note that $A\in (HEF)$ and in particular $AH$ is a diameter in $(AEHF).$ Therefore, $\angle OYA=\angle HYA = 90^{\circ}.$ Now $PQDX$ is cyclic iff $NMAY$ is cyclic. However, points $N,M,A,Y$ all lie on the circle with diameter $AO$ and we may conclude. Remark: This problem shares some similarities with IMO 2015/3.

Cute! WLOG $DEF$ be the orthic triangle, let $(AEF) \cap (AMN) = X^{*}$ where $M,N$ are midpoints of $AB,AC$, now by $\sqrt{-HA \cdot HD}$ inversion the queue points go to $M,N$ so it suffices to prove $O,H,X^{*}$ are collinear but this follows as $H,O$ are antipodes of $A$ in $(AEF)$ and $(AMN)$. Done.

Headsolved! Let $\omega$ be circle with diameter $AO$. Let $Y= OH \cap \omega$ other then $O$. Define $M,N$ be midpoint of $AB,AC$. Note that from $\angle XYA = \angle XDA = 90$ we have $A,Y,X,D$ cyclic. Do $-\sqrt{HA.HD}$ inversion about $H$. Note $P \leftrightarrow N$ , $ Q \leftrightarrow M$ and $ D \leftrightarrow A$. From fact that $A,Y,X,D$ cyclic we have $X \leftrightarrow Y$. Observe $A,M,N,Y,O$ are cyclic $\Leftrightarrow P,Q,X,D$ cyclic

This problem was proposed by Farid İsmayilov, Azerbaijan.

Let $BH\cap (ABC)=E,CH\cap (ABC)=F,LF\cap KE=R$ Let $K,L$ be the antipodes of $B,C$ on $(ABC)$ and $S$ be the foot of the altitude from $R$ to $OH$. Invert from $H$ with radius $\sqrt{-HB.HE}$. $B\leftrightarrow E,C\leftrightarrow F,P\leftrightarrow K,Q\leftrightarrow L,X\leftrightarrow S,D \leftrightarrow R$ We will prove that $RSKL$ is cyclic. Take $(ABC)$ unit circle. \[\frac{r+a+b+c}{2}=a\iff r=a-b-c\]\[s=\frac{\overline{(a+b+c)}(a-b-c)+(a+b+c)\overline{(a-b-c)}}{2\overline{(a+b+c)}}=\frac{(ab+bc+ca)(a-b-c)+(a+b+c)(bc-ab-ac)}{2(ab+bc+ca)}\]\[s=\frac{-ab^2-abc-ac^2}{(ab+bc+ca)}\]\[\frac{k-r}{k-s}.\frac{l-s}{l-r}=\frac{c-a}{-b+\frac{ab^2+abc+ac^2}{(ab+bc+ca)}}.\frac{-c+\frac{ab^2+abc+ac^2}{(ab+bc+ca)}}{b-a}\]\[=\frac{c-a}{b-a}.\frac{ab^2-bc^2}{ac^2-b^2c}=\frac{b(c-a)(ab-c^2)}{c(b-a)(ac-b^2)}\]\[\frac{b(c-a)(ab-c^2)}{c(b-a)(ac-b^2)}\overset{?}{=}\overline{(\frac{b(c-a)(ab-c^2)}{c(b-a)(ac-b^2)})}=\frac{(a-c)(c^2-ab)b}{(a-b)(b^2-ac)c}\]Which is true as desired.$\blacksquare$

Let $\Delta DEF$ be the orthic triangle of $\Delta ABC$. Take a negative inversion at $H$, swapping $A$ and $D$, $B$ and $E$, $C$ and $F$. Note that $P$ and $Q$ are the $B$-Queue point and the $C$-Queue point respectively, so $P'$ and $Q'$ are the midpoints of $AC$ and $AB$ respectively. The image of $(DPQ)$ is $(AP'Q')$, which has diameter $AO$. $X'$ lies on line $OH$ and circle $(AEF)$. It suffices to show that $X'$ also lies on the circle with diameter $AO$. But this is clear since $\angle AX'O = \angle AX'H = 90^\circ$.

After inversion with center $H$ and radius $\sqrt{-HA.HD}$, the new problem is :- In a triangle $\Delta ABC$, $M$ and $N$ are midpoints of $AB$ and $AC$ and $OH$ intersects $(AEF)$ again at $X$. To show that $X$ also lies on $(AMN)$. $\underline{Proof}$: Since $\angle AXH=90°= \angle AXO$; $X$ lies on the circle with diameter $AO$ which is $(AMN)$.

Let midpoints of sides $AB$ and $AC$ be $M$ and $N$. Let altitudes from $B$ and $C$ to $AC$ and $AB$ be $E$ and $F$. Consider $\sqrt{-HA \cdot HD}$ inversion. $P$ fixes to $N$ similarly $Q$ fixes to $M$. Thus, $(DPQ)$ fixes to $(ANM)$. Let $X'$ be inversion of $X$. We know that $AX'DX$ is cyclic which means $\angle AX'X=90$. This solves problem since $(AMNO)$ is circle with diameter $AO$ and $X-H-X'$ means $X'$ is on circle $(AMNO)$.