Let ABC be a triangle with circumcircle k. The points A1,B1, and C1 on k are the midpoints of arcs BC (not containing A), AC (not containing B), and AB (not containing C), respectively. The pairwise distinct points A2,B2, and C2 are chosen such that the quadrilaterals AB1A2C1,BA1B2C1, and CA1C2B1 are parallelograms. Prove that k and the circumcircle of triangle A2B2C2 have a common center.
Comment. Point A2 can also be defined as the reflection of A with respect to the midpoint of B1C1, and analogous definitions can be used for B2 and C2.
geometryJuniorBalkanshortlistparallelogramsconcentric