∴∠DAF+∠BAF=90，

∵AF⊥BE，

∴∠ABE+∠BAF=90，

∴∠ABE=∠DAF，

∵ In △ Abe and △DAF,

∠ABE=∠DAF，AB=AD，∠BAE=∠D

∴△ABE≌△DAF(ASA)，

∴af=be；

(2) Solution: MP and NQ are equal.

The reasons are as follows: As shown in the figure, intersection A is AF∨MP, CD is F, intersection B is BE∨NQ, and AD is E,? It is exactly the same as (1).

Prove; A is AF∨MP, CD is F, B is BE∨NQ, and AD is E.

In a square ABCD, AB=AD, ∠ BAE = ∠ d = 90,

∴∠DAF+∠BAF=90，

∵AF⊥BE，

∴∠ABE+∠BAF=90，

∴∠ABE=∠DAF，

∵ In △ Abe and △DAF,

∠ABE=∠DAF，AB=AD，∠BAE=∠D

∴△ABE≌△DAF(ASA)，

∴af=be；

Also: AFMP is a parallelogram and BNQE is a parallelogram.

∴AF=MP,NQ=BE

∴MP=NQ