You Mn//ab

∴ Mn: AC = BN/BC =1√ 2 (the square of similarity ratio is equal to the area ratio).

∴bn/nc = 1/(√2- 1)=√2+ 1

The intersection e is AC parallel lines ED, ED∑MN.

Because of the folding relationship

∴MN⊥BE，

∴BE⊥ED

That is, the delta bed is a right triangle,

∠∠BNM =∠ENM =∠ Ned =∠D

∴NE=ND=BN

So n is the midpoint of BD.

∵ Quadrilateral AEDC is a parallelogram, so CD = AE = X.

Let NC=a, then BN = (√ 2+1) a.

So nd = a+x = bn = (√ 2+1) a = > x/a =√2

∴AE:NC=√2