Let $ABC$ be an acute triangle with orthocentre $H$. Let $D$ be a point outside the circumcircle of triangle $ABC$ such that $\angle ABD=\angle DCA$. The reflection of $AB$ in $BD$ intersects $CD$ at $X$. The reflection of $AC$ in $CD$ intersects $BD$ at $Y$. The lines through $X$ and $Y$ perpendicular to $AC$ and $AB$, respectively, intersect at $P$. Prove that points $D$, $P$ and $H$ are collinear.
Problem
Source: Baltic Way 2024, Problem 13
Tags: geometry, geometric transformation, reflection, geometry proposed
math_comb01
17.11.2024 16:27
Let $BD \cap AC = J$, $CD \cap AB = K$, By angle chase we have $BXCY$ cyclic, and by reims we have $XY \parallel JK$, by homothety at $D$ it suffices to show that the perpendiculars from $J,K$ to $AB,AC$ are concurrent,call the intersection $W$, on the line $DH$, let the perpendiculars from $D$ to $AB,AC$ meet $AC,AB$ at $F,E$, then clearly $DEK \cup \{B\} \sim DFJ \cup \{C\}$ therefore $D,H$ and $W$ are collinear, as desired.
redred123
20.11.2024 13:57
how can $D$ be outside the circumcircle
Alex-Five
25.11.2024 00:03
Nice problem Let $E=XP\cap BD$, $F=YP\cap CD$. First, we show that $BCXY$ is cyclic. Indeed, follows as $\angle YBX = \angle ABD = \angle ACD = \angle XCY$. Next, we show that $EFXY$ is cyclic. Indeed, follows as $\angle EYF= 90^\circ - \angle YBA = 90^\circ - \angle XCA = \angle EXF $ Finally, we show that $BC$ is parallel to $EF$. This follows as $\angle DEF = 180^\circ - \angle YEF = \angle YXF = \angle YXC = 180^\circ - \angle YBC = \angle DBC$. We finish by homothety as $\bigtriangleup EPF $ and $\bigtriangleup BHC$ have pairwise parallel sides and are therefore homothetic. Therefore $BE$, $HP$ and $CF$ concur.
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