∫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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