2、b(n,0),a(m,0),bc=vn^2+ 1,ac=vm^2+ 1

ab=m-n,ab^2=m^2+n^2-2mn,mn=- 1,ab^2=m^2+n^2+2

bc^2+ac^2=n^2+ 1+m^2+ 1+m^2+n^2+2=ab^2

∠ABC=90 degrees

3. In three cases, when BC=AC, n 2+4n 2 = 4+4n 2, n=2 or -2, n = 2, n=-2.

AB=AC, (2-n) 2 = 4+4n 2, n=-3/4 or n=0 n=0 shed, n =-3/4.

BC=AB，n^2+4n^2=(2-n)^2，x=(- 1 √5)/2

1, it is proved that folding = & gt∠ABE=∠MBE= 1/2∠ABM, ∠NDF=∠CDF= 1/2∠NDC.

ABCD is rectangular, BD is diagonal = & gt∠ABM=∠NDC, ∠DBC =∠ ADB.

∠MBE=∠NDF，∠DBC =∠ADB = & gt； ∠MBE+∠DBC =∠ADB+∠NDF = & gt； ∠EBC=∠ADF

∠MBE=∠NDF，∠DBC =∠ADB = & gt； 180 degrees -∠MBE-∠ADB= 180 degrees -∠NDF-∠DBC

=>∠ Bed =∠BFD

∠BED=∠BFD, ∠EBC=∠ADF, and two sets of quadrangles with equal diagonals are parallelograms.

= & gtBFDE is a parallelogram.

2.Diamond = & gt∠CBD =∠EBD =∠ABE = 30 = & gt； AE = 2/V3 = & gt； DE=BE=4/V3

=> area =4/V3*2=8V3/3