Extend DA to point e, make AE=CN, and connect BE.

∫∠ bad +∠C= 180.

∴∠EAB=∠C.

AB = BC，AE=CN，

∴△ABE≌△CBN.

∴∠EBA=∠CBN,BE=BN.

∴∠EBN=∠ABC.

∠∠ABC = 80，∠MBN=40，

∴∠EBM=∠NBM=40。

BM = BM，

∴△EBM≌△NBM.

∴EM=NM.

∴MN=AM+CN.

(2)

MN & ltAM+CN

(3)

Take M and N from AD and BC, extend DC to G, make CG=AM, and connect BG, BM, BN and MN.

Let MD = x and nd = y.

The side length of a square ABCD is 1.

∴ab=bc=ad=cd= 1,∠a=∠bcg=90

∴am= 1-x,cn= 1-y,,△bam≌△bcg(sas)

∴CG=AM= 1-y,BM=BG

∴gn=cn+cg=( 1-x)+( 1-y)=2-x-y

∫△MDN circumference is 2.

∴MN=2-x-y

∴MN=GN

BM = BG，，BN=BN

∴△BMN≌△BGN(SSS)

∴S△BMN

=S△BGN

=S△BCN+ S△BCG

= S△BCN+S△ bam

∴S square ABCD

= S△BCN+S△ bam+S△BMN+S△MDN

=2 S△BMN+S△MDN

∴S△BMN= (squared ABCD- S△MDN)/2

∫S squared ABCD =1*1=1,and find the minimum value of S△BMN.

∴S△MDN should take the maximum value.

∴DM=DN

That is x = y.

∴MN=x*√2

∴C△MDN=x+x+ x*√2=2

X=2-√2。

∴S△MBN

= S squared ABCD-S△BCN-S△ bam -S△MDN

= 1-[( 1-x)* 1]/2-[( 1-x)* 1]/2-(x * x)/2

=- 1+√2