Problem

Source: All-Russian MO 2023 Final stage 11.6

Tags: geometry, 3D geometry, tetrahedron



The plane $\alpha$ intersects the edges $AB$, $BC$, $CD$ and $DA$ of the tetrahedron $ABCD$ at points $X, Y, Z$ and $T$, respectively. It turned out, that points $Y$ and $T$ lie on a circle $\omega$ constructed with segment $XZ$ as the diameter. Point $P$ is marked in the plane $\alpha$ so that the lines $P Y$ and $P T$ are tangent to the circle $\omega$.Prove that the midpoints of the edges are $AB$, $BC$, $CD,$ $DA$ and the point $P$ lie in the same plane.