∵OP=1/2AB=1/2AC,∴OP‖AC,又∵PD⊥AC,∴PD⊥OP.
连接AP,△ABP为直角△,AP⊥BC,∵∠CAB=120°,∴∠ABP=30°
∵AB=2,∴BP=√3.
又∵OP‖AC,∴BP=PC,BP=2BP=2√3 .