Because △BMN is folded in half by △ABM, so

△BmN △ ABM, so △ BMN is also a right triangle, BN⊥MN, ∠ABM=∠MBN.

EF is the polyline of rectangular ABCD, that is, AE=EB, DF=FC,

∠DFE=∠CFE, EF⊥DC, ∴EF is parallel to AD and BC,

NF is ∴MN=NP. The center line of trapezoidal MDCP.

Right triangle BNM is equal to triangle BNP, so

∠MBN=∠PBN=∠ABM，

∠MBN+∠PBN+∠ABM=90，

∴∠MBN=∠PBN=∠ABM=30

∠MBP=∠MBN+∠PBN=60

∠BMP=∠BPM=60

So delta △BMP is an equilateral triangle.

1. From the first folding, we can know that n is the midpoint of MP.

2. According to the second folding, ∠MNP is a right angle, so BN is the middle vertical line of MP, so BM=BP, ∠MBN=∠PBN.

3. From the second folding, we can know that ∠ABM=∠MBN.

4. According to 2 and 3, ∠ABM=∠MBN=∠PBN=300, so ∠MBP=600 and △BMP is an equilateral triangle, because BM=BP.

First, the angle BNM is formed by folding the angle A, so it is a right angle, and the point is the midpoint of MP. So we can know that the triangle BMP is isosceles (BP=BM), and the vertical line in the middle bisects the vertex angle, so ∠MBN=∠PBN, and because it is folded, ∠ABM =∠MBN =∞.