∫CE:AC = 2:3，AE= AC-CE

∴AE:AC= 1:3

∫AD:DB = 1:2，AB=AD+BD

∴AD:AB= 1:3

∴DE∥BC, and DE: BC = 1:3 (△ADE and △ABC are similar, and the similarity ratio is 1:3).

M and n are the midpoint of the edge of BE and CD, respectively.

And DE∑BC is known.

∴MN∥DE∥BC

∴ Quadrilateral DBCE is a trapezoid.

(2)

If MN crosses AC to F, MF∨DE∨BC and F must be the midpoint of EC.

∴MF=BC/2

NF=DE/2

MN = MF-NF = BC/2-DE/2 = 3DE/2-DE/2 = DE

∴MN:DE= 1: 1

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