Problem 4.
Let triangle ABC satisfy AB < AC < BC. Let $I$ and $\omega$ be the incenter and the incircle of triangle $ABC$, respectively. Let $X$ be a point on line $BC$, different from $C$, such that the line through $X$ and parallel to $AC$ is tangent to $\omega$. Similarly, let $Y$ be a point on line $BC$, different from $B$, such that the line through $Y$ and parallel to $AB$ is tangent to $\omega$. Line $AI$ intersects the circumcircle of triangle ABC at P ≠ A. Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.
Prove that ∠KIL + ∠YPX = 180°.
Proposed by Dominik Burek, Poland