1 证明：连接AC,AN,BN∵PA⊥平面ABCD ∴PA⊥AC,PA⊥AB,既△PAC,△PAB均是Rr△∵N是PC的中点∴NA是直角三角形PAC的中线∴NA=（1/2）PC∵PB^2=AB^2+PA^2 PC^2=PA^2+AC^2=PA^2+AB^2+BC^2=PB^2+BC^2∴△PBC是Rr△∵BN是...