(1) ne = MB and NE⊥MB.

(2) established.

Proof: Connect AE and extend the intersection of NE and BM at point F. 。

∫E is the midpoint of CD, AB = BC = 12cd,

Ab = European community.

And ∵AB∨CD,

That is AB∨? CE。

A quadrilateral is a parallelogram.

∫∠C = 90，

∴ Quadrilateral ABCE is a rectangle.

ab = BC，

A quadrilateral is a square.

∴ AE=AB。

In the isosceles right triangle AMN,

∴ AN=AM，∠NAM=90。

∴ ∠ 1+∠2=90 .

∵∠ 2+∠ 3 = 90,

∴ ∠ 1=∠3.

∴△nae?△mab。

∴ NE=MB，∠AEN=∠ABM。

∴ ∠4=∠6.

∵ ∠5=∠6,

∴ ∠4=∠5.

∠∠EMF =∠BMC，

∴ ∠EFB=∠C=90。

∴ BM⊥NE.