2024 IMO Shortlist G2

Let $ABC$ be a triangle with $AB<AC<BC$, incentre $I$ and incircle $\omega$. Let $X$ be the point in the interior of side $BC$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be the point in the interior of side $BC$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ again at $P\ne A$. Let $K$ and $L$ be the midpoints of $AB$ and $AC$, respectively.

Prove that $\angle KIL+\angle YPX=180^\circ$.

设 $ABC$ 为三角形，满足 $AB<AC<BC$，其内心为 $I$，内切圆为 $\omega$。设 $X$ 为边 $BC$ 内部一点，使得过 $X$ 且平行于 $AC$ 的直线与 $\omega$ 相切。类似地，设 $Y$ 为边 $BC$ 内部一点，使得过 $Y$ 且平行于 $AB$ 的直线与 $\omega$ 相切。设 $AI$ 与三角形 $ABC$ 的外接圆再次交于 $P\ne A$。设 $K,L$ 分别为 $AB,AC$ 的中点。

证明 $\angle KIL+\angle YPX=180^\circ$。