Given a triangle $PQR$, the inscribed circle $\omega$ which touches the sides $QR, RP$ and $PQ$ at points $A, B$ and $C$, respectively, and $AB^2 + AC^2 = 2BC^2$. Prove that the point of intersection of the segments $PA, QB$ and $RC$, the center of the circle $\omega$, the point of intersection of the medians of the triangle $ABC$, the point $A$ and the midpoints of the segments $AC$ and $AB$ lie on one circle.
Problem
Source: 2013 XVI All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p9
Tags: geometry, Concyclic, Ukrainian TYM