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=(S square 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