Since ( AC ) is a diameter, step 2’s circle has center ( A ) radius ( 2R ). The chord ( PQ ) is the radical axis of ( \omega ) and ( \Gamma ), which lies on the perpendicular bisector of ( AC ), hence passes through ( O ). Therefore ( BD ) is the diameter perpendicular to ( AC ), forming a square with vertices at the endpoints of these perpendicular diameters.
Draw line ( OA ), get ( C ) (other intersection). 2E: Draw circle with center ( A ) through ( C ). This circle intersects the original circle at points ( P ) and ( Q ). 3E: Draw line ( PQ ). Where is ( PQ )? It is the perpendicular bisector of ( AC ) and passes through ( O ). So ( PQ ) is the perpendicular line to ( OA ) through ( O ). euclidea 2.8 3e
Circle ( \omega ) with center ( O ) and point ( A ) on ( \omega ). Goal: Construct all vertices of a square inscribed in ( \omega ) using exactly 3 elementary moves. Since ( AC ) is a diameter, step