Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.5 National olympiad Prove it

Let $A B C$ be an acute triangle with orthocentre $H$ . Let $D$ be a point outside the circumcircle of triangle $A B C$ such that $\angle A B D=\angle D C A$ . The reflection of $A B$ in $B D$ intersects $C D$ at $X$ . The reflection of $A C$ in $C D$ intersects $B D$ at $Y$ . The lines through $X$ and $Y$ perpendicular to $A C$ and $A B$ , respectively, intersect at $P$ . Prove that points $D, P$ and $H$ are collinear.

Solution

From the reflections, we have

$\angle D B X=180^{\circ}-\angle D B A=180^{\circ}-\angle D C A=\angle D C Y$

(Fig. 15), so points $B, C, X, Y$ are concyclic.
Define $Q=X P \cap B D$ and $R=Y P \cap C D$ (Fig. 16). Then due to the right angles, we find $\angle D Y R=\angle D X Q$ . Hence points $Q, R, X, Y$ are concyclic, too.
Consequently, $\angle D Q R=\angle D X Y=\angle D B C$ , so $B C \| Q R$ . Since also $B H \| P Q$ and $C H \| P R$ , it follows that triangle $B H C$ is a homothetic image of triangle $Q P R$ with center $D$ . Hence $D, P$ and $H$ are collinear.
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Figure 15
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Figure 16

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.