Let AB be a chord and angleAOB be the angle subtended by it at centre and angleAPB be the angle subtended at any point P on the remaining part of circle. Now join PO and extend it to some point Q. angleAPQ is external angle to triangle POA.
→ angleAPQ= angleOPA + angle OAP.
Since OP and OA are radii of circle and hence equal, the triangle OAP is isoceles triangle. Therefore angle OPA = angle OAP. → angle APQ =
2×angleOPA. ....(I)
Similarly considering triangle OPB, angle BPQ =
2× angle OPB....(ii)
Adding (i) and (ii), angle APQ + angle BPQ =
2× (angle OPA + angle OPB)
angle AOB =
2× angle APB ------ PROVED