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Jinshan District Ermo 20 16 Mathematics
(1) Proof: ∵ab∨CD, AC⊥CD.

∴∠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.