Let AB be any chord passing through the fixed point P. Draw the diameter AOC, join CP and let it meet the circumference at D Then in triangle ACP, AC² = AP² + CP² together with either 2AP. PB or 2 CP. PD according as CP is projected on AP or AP on CP. Hence AP.PB = CP.PD. Now draw the diameter DOE, join EP and let it meet the circumference at F. Then the theorem is true for chords CD. EF.