证明：作△ABC的外接圆⊙O，并作OE⊥AB于E，OF⊥AC于F，连接OP、OQ、OB、OA，∵O是△ABC的外心，∴OE=OF，OB=OA，由勾股定理得：BE2=OB2-OE2，AF2=OA2-OF2，∴BE=AF，∵AP=BQ，∴PF=QE，∵OE⊥AB，OF⊥AC∴∠OFP=∠O...