在棱AB上取一点P,使AP=2AB/3,连结PN、PM,
∵棱长AB=BC=AA1=a,
∴AC=√2a,A1B=√2a,
∵A1M=AN=√2a/3,
∴A1M=2A1B/3,AN=2AC/3,
∵AP/AB=AN/AC=2/3,
∴PN//BC,（三角形平行比例线段定理逆定理）,
同理,PM//AA1,
∵AA1//BB1,
∴PM//BB1,
∵PN∩PM=P,
BC∩BB1=B,
∴平面PMN//平面BB1C1C,
∵MN∈平面PMN,
∴MN//平面BB1C1C.
2、∵PN//BC,
∴PN/BC=AN/AC=2/3,
∴PN=2BC/3=2a/3,
同理,PM=2AA1/2=2a/3,
∵AA1⊥平面ABCD,PM//AA1,
∴AA1⊥平面ABCD,
∵PN∈平面ABCD,
∴PM⊥PN,即∵PM=PN=2a/3,
∴△MPN是等腰RT△,
∴MN=√2PM=2√2a/3.