∫∠B = 60，AB= 12，

∴sin60 =AE 12，

∴AE=6√3，

∴BE=6, which can also prove that FC=6.

∴bc=be+ef+fc=6+4+6= 16；

(2) Delta △PBM with high PG,

∫ The area of the isosceles trapezoid ABCD is:12 (ad+BC)? AE = 1 2×(4+ 16)×6√3 = 60√3

∫PM bisects the area of trapezoidal ABCD,

∴S△PBM=30√3，

∫BM = 12，

∴PG=5√3，

∫∠B = 60，

∴PB=5√3/sin60，

∴pb= 10；

(3) When m is on BC, the circumference of trapezoidal ABCD is 4+12+16+12 = 44.

∫PB = 10, BM= 12, PB+BM=22 (in line with the meaning of the question),

PB= 12, BM= 10 PB+BM=22.

There is a straight line pm that fits the meaning of the question.