ÖMO 2018 Problem 4

Let $ABC$ be a triangle and $P$ a point inside the triangle such that the centers $M_B$ and $M_A$ of the circumcircles $k_B$ and $k_A$ of triangles $ACP$ and $BCP$ , respectively, lie outside the triangle $ABC$ . In addition, we assume that the three points $A, P$ and $M_A$ are collinear as well as the three points $B, P$ and $M_B$ . The line through $P$ parallel to side $AB$ intersects circles $k_A$ and $k_B$ in points $D$ and $E$ , respectively, where $D, E \ne P$ . Show that $DE = AC + BC$ .

(Proposed by Walther Janous)