Let $A$, $B$, and $C$ be three points such that $B$ is the midpoint of segment $AC$ and let $P$ be a point such that $\angle PBC=60$. Equilateral triangle $PCQ$ is constructed such that $B$ and $Q$ are on different half=planes with respect to $PC$, and the equilateral triangle $APR$ is constructed in such a way that $B$ and $R$ are in the same half-plane with respect to $AP$. Let $X$ be the point of intersection of the lines $BQ$ and $PC$, and let $Y$ be the point of intersection of the lines $BR$ and $AP$. Prove that $XY$ and $AC$ are parallel.

Let $O$ be the center of the equilateral $\triangle APR$ and $M$ be the midpoint of $AP$. Because $\angle ABP = \angle AOP = 120^\circ$, we know that $B \in \odot(POA)$ and since $RA,RP$ are tangents to $\odot(POA)$, we see that $BY$ is the $B$-symmedian of $\triangle PBA$ which is isogonal of the median $BM$. This implies that

$\angle PBY=\angle ABM=\angle PCA$. However, $\triangle PAC \cong \triangle PRQ$ (directly), so $\angle PQR=\angle PCA$ $\Longrightarrow$ $\angle PQR=\angle PBY$ $\Longrightarrow$ $PQRB$ is cyclic $\Longrightarrow$ $\angle RBQ=\angle RPQ=\angle APC$ $\Longrightarrow$ $PXBY$ is cyclic $\Longrightarrow$ $\angle PYX=\angle PBX=\angle PRQ=\angle PAC$ $\Longrightarrow$ $XY \parallel AC$, as desired. $\square$
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