Let $AHKO$ be a parallelogram. Then $APKQ$ is also a parallelogram. Notably, let $M$ be the midpoint of $BC$, and $M_B$ the midpoint of $AB$; then $MM_B$ is parallel to $KP$. If $P'$ is the reflection of $P$ across $M_B$, then $OP'$ is parallel to $KP$; thus $\angle{OPA} = \angle{OP'B} = \angle{A}$. Now, since $AQ = PK$, we find that $X$ is just the reflection of $K$ across $AB$. After this, the problem is just some vector calculations. Note that $K$ is the circumcenter of $(BHC)$; suppose $(BHC)$ intersects $AB$ again at $B'$, $AC$ again at $C'$. Then we want to show $H, K, X + Y - O$ are collinear. Since $X = B + B' - K$, $Y = C + C' - K$, we find that it suffices to show that $H, K, B' + C' - K$ are collinear. But $HB' = HA = HC'$ and $KB' = KC'$, so all three points are on the perpendicular bisector of $B'C'$ as desired.

Denote $D$ and $M$ by the midpoint of $BC,XY$,respectively. First, we recall some basic lemmas involving nine point center. Lemma 1 : Let $A’$ be the reflection of $A$ across $O_9$ then, $A’$ is the reflection of $O$ across $BC$. Proof : It is well known that the radius of nine point circle is half the radius of the circumcircle. $\implies O_9D = \frac{AO}{2}$ which means $\frac{O_9D}{AO}=\frac{A’O_9}{A’A}$ It is sufficient to prove that $O_9M \parallel AO $ which can be done by angle chasing. $\qquad \square$ Lemma 2 : Let $K$ be the circumcenter of $\triangle BOC$ then, $AO_9$ and $AK$ are isogonal conjugate wrt. $\angle BAC$ Proof Perform an inversion at $A$ radius $\sqrt{\frac{bc}{2}}$ then reflect the image across the angle bisector of $\angle BAC$ We can easily see that $(BOC) \mapsto$ nine point circle, their center must be isogonal conjugate wrt. $\angle BAC$ $\qquad \square$ We’re now ready to tackle the problem. Step 1 (parallelogram trick) Observe that the condition $O_9$ is the midpoint of $PQ$ motivates us to apply the parallelogram trick Construct parallelogram $APA’Q$ which means that $A’$ is reflection of $A$ across $O_9$. By lemma 1, $A’$ is the reflection of $O$ across $BC$. Consider $\angle A’OC = \frac{\angle BOC}{2}= \angle BAC = \angle A’QC$ $\implies$ $A’,O,Q,C$ are concyclic Similarly, we can easily show that $A’,O,P,B$ are concyclic. So, we obtain $\angle OQA=180^\circ - \angle OQC = \angle OA’C = \angle BAC $ We can prove in the same way that $\angle OPA = \angle BAC$. Step 2 (spot similar triangles) Since $AQ=XP, QY=PA$ and $\angle AQY= \angle XPA = 180^\circ - \angle BAC$$\implies \triangle AQY \cong \triangle XPA$ Moreover, notice that $\angle APA’ = \angle YQA = 180^\circ - \angle BAC, AP=YQ$ and $PA’=QA$ $\implies \triangle A’PA \cong \triangle AQY $ Similarly, $\triangle AQA’ \cong \triangle XPA \implies $ $\triangle AQY ,\triangle XPA , \triangle AQA’$ and $\triangle A’PA $ are all congruent. Consider isosceles triangle $AXY$ and $OBC$. By trivial angle chasing we can see that $\angle XAY = \angle BOC = 2 \angle BAC \implies \triangle AXY \sim \triangle OBC$ Step 3 (Find the spiral center) We have that $AP$ bisects $\angle XAO_9$, $AQ$ bisects $\angle YAO_9$ and $AM$ bisects $XAY$ By trivial angle chasing, we conclude that $AO_9$ and $AM$ are isogonal conjugate wrt. $\angle BAC$ Let $AM$ intersects $OA’$ at $T$. By lemma 2, we conclude that $T$ is the circumcenter of $\triangle BOC$ We now claim that $T$ is the spiral center which sends $\triangle AXY$ to $\triangle OBC$ Proof We can just show that $T$ is spiral center which sends $XA$ to $BO$ which is equivalent to showing $\triangle TBX \sim \triangle TOA$ We already have that $TB=TO$ and $BX=OA$. We’re left to show that $\angle TBX = \angle TOA$ Consider $$\angle TBX = 360^\circ - \angle XBA - \angle ABO - \angle OBT = 360^\circ - \angle ABA’ - (90^\circ - \angle ACB) - \angle BAC = 360^\circ - (\angle ABC + 90^\circ - \angle BAC ) - (90^\circ - \angle ACB) - \angle BAC = 2\angle ACB + \angle BAC = \angle TOA$$Step 4 (Finish) Denote $\phi$ by a spiral similarity centered at $T$ which maps $\triangle AXY \mapsto \triangle OBC$ It follows that $\phi$ maps $XB \mapsto AO \mapsto CY$ $\implies \triangle TBX \sim \triangle TOA \sim \triangle TCY \qquad (\spadesuit)$ Since $D$ and $M$ are midpoints of $BC$ and $XY$, respectively. $\implies \triangle TBX \sim \triangle TDM \sim \triangle TCY \qquad (\clubsuit)$ Combine $(\spadesuit),(\clubsuit)$, we conclude that $\triangle TOA \sim \triangle TDM$ Since $T,D,O$ are collinear, we conclude that $AO \parallel MD$ Recall the fact that $O_9D \parallel AO$ $\implies O_9,M,D$ are collinear $\qquad \blacksquare$