∴∠BAP=∠DCP=90，

E and f are the midpoints of line segments BP and DP, respectively,

∴AE=PE=BE,CF=PF，

∴∠EAP=∠EPA,∠CPF=∠FCP，

∠∠EPA =∠CPF，

∴∠EAP=∠FCP，

∴ae∥cf；

(2) proof: connect EM, EN,

∵M and e are the midpoint of AP and BP respectively,

∴EM∥AB，

∫AB∨CD，

∴ME∥DC, that is, em∨cn.

∫AB∨CD，

∴△AEB∽△QED，

∴AEEQ=BEDE，

AE = BE，

∴DE=EQ，

∫N is the midpoint of DQ,

∴EN⊥AQ，

∫∠ACD = 90，

∴EN∥MC，

The quadrilateral McKern is rectangular,

∴MN=CE.