Question 55
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 $D, E \neq P$.
Show that $DE = AC + BC$.
Show that $DE = AC + BC$.