(1) Calculate AD∨BC according to the properties of rectangle, and deduce OM=ON according to OB=OD and AD∨BC to get parallelogram BMDN, and deduce rhombic BMDN; ?

(2) Find DM=BM according to the diamond property, and in Rt△AMB, find BM 2 = AM 2+AB 2 according to Pythagorean theorem, and deduce that X 2 = X 2 ﹣16x+64+16.

Answer:?

(1) Prove: ∵ Quadrilateral ABCD is a rectangle,?

∴AD∥BC,∠A=90，

∵MN is the middle vertical line of BD,

∴BM=DM，

OB = OD，

∴ quadrilateral BMDN is a parallelogram.

∵MN⊥BD，

The parallelogram BMDN is a diamond.

(2) Solution: ∫ The quadrilateral BMDN is a diamond,

∴MB=MD？

Let MD be x, then MB=DM=x,?

In Rt△AMB, BM 2 = am 2+ab 2.

X 2 = (8 ﹣ x) 2+4 2，？

Solution: x=5,?

A: MD is 5.