解: (1) ∵AB=AC,∠BAC=60°, ∴△ABC是等边三角形, ∵∠APB+∠ACB=180°, ∴∠APB=120°; (2)当点P运动到的中点时,PD⊥AB,如图1,连接PC,OA,OB, 设⊙O的半径为r,则CP=2r, 又∵⊙O为等边△ABC的外接圆, ∴∠OAB=30°,在Rt△OAD中, ∵OD=OA=, ∴CD=+r=, ∴CD:CP=:2r=3:4; (3)PC=AP+PB 证明:如图2,在AP的延长线上取点Q,使PQ=PB,连接BQ, ∵∠APB=120°, ∴∠BPQ=60°, ∴△BPQ是等边三角形, ∴PB=BQ, ∵∠CBP=∠CAB+∠ABP=60°+∠ABP,∠ABQ=∠QBP+∠ABP=60°+∠ABP, ∴∠ABQ=∠CBP,在△ABQ和△CBP中,PB=QB,∠CBP=∠ABQ,CB=AB, ∴△ABQ ≌ △CBP, ∴CP=AQ=AP+PQ=AP+PB, 即PC=AP+PB; |