Fold the rectangular paper ABCD in half, so that AD and BC overlap, and get the crease EF.

So EF∨AD∨BC

△ NBM obtained by folding△ △ABM along BM.

∴△ABM≌△NBM

∴∠ABM=∠MBN and ∠ MNB = 90, namely BN ∠ MP.

EF∨AD∨BC and AE = DE

∴MN = NP

(Three parallel lines cut two straight lines, and the corresponding line segments are proportional. )

BN is an ordinary edge.

∴△BMN≌△BPN

∴∠MBN=∠PBN and BM = BP.

∴∠ABM=∠MBN = ∠PBN

∠ABC = 90°

∴∠ABM=∠MBN = ∠PBN = 30

∴ ∠MBP = 60

BM = BP

△BMP is an equilateral triangle