China-TST-A

Geometry Difficulty 8.4 Shortlist Prove it China

Let point $P$ lie on the nine-point circle of triangle $ABC$ . A line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$ . A line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$ . Let $H$ be the orthocenter of triangle $ABC$ , and let $D$ and $M$ be the midpoints of segments $BC$ and $AQ$ , respectively. Prove that $XH \perp DM$ .

Solution

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 $\triangle DPQ \sim \triangle NPA$ . Therefore, $\frac{DQ}{NA} = \frac{PQ}{PA}$ .

Since $XA \perp AQ$ , we have $\triangle AQP \sim \triangle XAP$ , 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 $\triangle XAN \sim \triangle AQD$ , 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 $\triangle XHA \sim \triangle MDQ$ , which proves that $DM \perp XH$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.