∵ABCD is a rectangle,

∴AM∥DN，

∴∠KNM=∠ 1.

∫∠KMN =∠ 1，

∴∠KNM=∠KMN.

∴△MNK is an isosceles triangle.

(2) There are two situations:

Case 1: Fold the rectangular piece of paper in half so that point B coincides with point D, and point K also coincides with point D at this time.

Let MK=MD=x, then AM = 25-X. In Rt△DNM, we can get from Pythagorean theorem.

,?

Solve,.

That is MD = nd = 13.

∴S △MNK =32.5。

Case 2: Fold the rectangular paper in half along the diagonal AC, and the crease is AC.

Let MK="AK=" CK=x, then DK=25-x, which can also be obtained.

That is MK = NK = 13.

∴S △MNK =32.5。

Case 1: Fold the rectangular piece of paper in half so that point B and point D coincide, and point K also coincides with point D at this time; Case 2: Fold the rectangular paper in half along the diagonal AC, and the crease is AC.