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