China-TST-2025A

Proof: Let $N$ be the midpoint of $AH$. By the properties of the nine-point circle, $DN$ is its diameter, so $DP\perp PN$. Since $AH\perp BC$ and $AP\perp PQ$, we have $△DPQ\sim △NPA$. Therefore, $\frac{DQ}{NA}=\frac{PQ}{PA}$.

Since $XA\perp AQ$, we have $△AQP\sim △XAP$, which gives $\frac{PQ}{PA}=\frac{AQ}{XA}$. Combining these two results yields $\frac{DQ}{NA}=\frac{AQ}{XA}$.

Moreover, we observe that: $$\angle XAN=90^{\circ }-\angle NAQ=\angle AQD$$ This implies that $△XAN\sim △AQD$, and consequently: $$XA\cdot DQ=AN\cdot AQ=AH\cdot QM$$ Finally, noting that: $$\angle XAH=90^{\circ }-\angle NAQ=\angle MQD$$ we conclude that $△XHA\sim △MDQ$, which proves that $DM\perp XH$. $\square$