As shown in figure (3), make a chord BE⊥CD at point B, connect AE and CD at point P, and connect OB, OE, OA and PB.

∵BE⊥CD，

∴CD bisects BE, that is, point E and point B are symmetrical about CD.

∵AC？ It's 60 degrees, point b is the midpoint of AC,

∴∠BOC=30，∠AOC=60，

∴∠EOC=30，

∴∠AOE=60 +30 =90，

∫OA = OE = 1，

∴AE=2OA=√2

The length of AE is the minimum value of BP+AP. ∴BP+AP=√2

(3) Expansion and extension

As shown in Figure (4), that is, point P is the symmetrical point E about straight line AB, point P is the symmetrical point F about straight line BC, and EF is connected to points M and N of AB and BC respectively.

Hope to adopt thank you!