\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Shortlist 2005 G3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 3}
\maketitle
\section*{Problem}
Let $ABCD$ be a parallelogram.
A variable line $g$ through the vertex $A$ intersects
the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively.
Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$.
Show that the measure of angle $\angle KCL$ is independent of the line $g$.
\section*{External Link}
\url{https://aops.com/community/p512720}
\newpage
\section*{Solution}
Here is a synthetic solution.
Let $\alpha = \angle BAK = \angle KAX$ and $\beta = \angle XAL = \angle LAD$.
Since $\angle ABK = \angle ADL = 180\dg-(\alpha+\beta)$,
it follows that
\[ \triangle ABK \sim \triangle LDA \implies
\frac{AB}{LD} = \frac{BK}{DA}
\implies \frac{CD}{LD} = \frac{BK}{BC}
\implies \triangle LDC \sim \triangle CBK. \]
Then, $\angle BCK + \angle LCD = 180\dg-(\alpha+\beta)$,
which is fixed.
On the other hand, $\angle BCD = 2(\alpha+\beta)$ is fixed too.
Therefore $\angle KCL$ is fixed as desired.
Moving points should also work; see
\url{https://aops.com/community/c6h87927p8743952} for example.
\end{document}