∴∠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.