∠OAP=∠OBP=90°,\angle OAP=\angle OBP=90\degree,∠OAP=∠OBP=90°,
OA=OBOA=OBOA=OB like radiuses, OP is common side,
so △OAP=△OBP.\triangle OAP=\triangle OBP.△OAP=△OBP.
So AP=BPAP=BPAP=BP and ∠APO=∠BPO.\angle APO=\angle BPO.∠APO=∠BPO.
Line segment PA is congruent to line segment PB and line segment PO bisects angle APB.
Q.E.D.
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