(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.