Idea 1: To prove that DM = MN, we can consider proving that the two triangles where DM and MN are located are congruent. In the figure, MN is at an obtuse angle △MBN and DM is at a right triangle, so we need to construct an obtuse triangle. So there is the following solution 1.

The solution 1 conjecture DM = Mn is proved as follows:

As shown in figure 3, ah = am is intercepted at the edge of AD and can be obtained by squaring ABCD and BECF.

AD=AB，∠A=∠ABC=90，∠CBF=45，

Then DH = MB,

∠DHM=∠MBN= 135 .

∵DM⊥MN,∴∠AMD+∠ 1=90，

∫∠AMD+∠2 = 90，

∴∠ 1=∠2,

∴△DHM≌△MBN,∴DM=MN。

Idea 2: As mentioned above, the right triangle where MN is located can also be constructed to be congruent with Rt△ADM where DM is located. So there is the following solution 2.

Solution 2 conjectures DM = Mn, which is proved as follows:

As shown in Figure 4, if N is passed, it is NG⊥BE and the vertical foot is G. ..

Then ∠ mgn = ∠ a = 90.

∵DM⊥MN,∴∠AMD+∠ 1=90 .

∫∠AMD+∠2 = 90，

∴∠ 1=∠2,

∴Rt△ADM∽Rt△GMN，

∴ =?

Let ad = a, BM = x, ng = y,

If nbg = 45, BG = ng = y (a > x).

∴ = = = 1,

∴ a = x+y, that is, ad = mg.

∴△DHM≌△MBN,∴DM=MN。

(if a = x, DM = ad, Mn = ab, DM = Mn still holds. )

Idea 3: To prove DM = Mn, we can also construct an isosceles triangle with this as the two waists, and the conclusion will be proved. However, it is not easy to prove directly connecting DN, so we can consider turning the problem into proving that DM and MN are equal to the third line segment. So there is solution 3.

Solution 3 conjectures DM = Mn, which is proved as follows:

As shown in fig. 5, let △MBN be an axisymmetric figure △MBP about a straight line AE, and connect DB and MN at point G, then there is:

△MBN?△MBP，∠2=∠ 1=45，

∠P=∠MNB，MN=MP .

Pass the square of ABCD and BEFC, and then

∠MBN=∠DBC+∠CBN=45 +45 =90，

∴∠dbp=∠dbn+∠ 1+∠2= 180

∴D, B and P are on the same straight line.

And ∵DM⊥MN, ∠ DGM = ∠ NGB,

∴∠3=∠MNB,∴∠3=∠P，

∴DM=MP,∴DM=MN。