解:(1)∵△ABC为等边三角形,∴∠BAC=60°, ∵点P为等边△ABC外接圆劣弧BC上一点,∴∠BPC+∠BAC=180°,∴∠BPC=120°, (2)在PA上截取PD=PC, ∵AB=AC=BC,∴∠APB=∠APC=60°,∴△PCD为等边三角形,∴∠ADC=120°, ∴△ACD≌△BCP,∴AD=PB,∴PA=PB+PC; (3)∵△CDM∽△ACM,∴CM:AM=DM:MC=DC:AC=2:4=1:2, 设DM=x,则CM=2x,BM=4-2x,PM=2-x,AM=4x,∵△BPM∽△ACM,∴BP:AC=PM:CM, 即3x:4=(2-x):2x,解得,x=(舍去负号),则x=,∴CM= . |