Analysis: 1) Calculate ∠ A = ∠ C = 45 according to the properties of isosceles right triangle, and calculate ∠ AFP = ∠ C = 45 according to the fact that two straight lines are parallel and the congruence angle is equal, thus judging that △APF is an isosceles right triangle;

(2)① Let AP=CQ=x represent PB, and then calculate the solution according to the triangle area formula;

② If Q passes, QF⊥AC passes through the AC extension line at F, and the congruence of △QCF and △PAE is proved by "corner edge". According to the equivalence of corresponding edges in congruent triangles, AE=CF and EP=QF can be obtained, thus AC=EF can be obtained, and then the congruence of △EPD and △FQD can be proved by "corner edge".

Solution: (1) Solution: ∵∠ B = 90, AB=BC,

∴∠A=∠C=45，

∫PF∨BC，

∴∠AFP=∠C=45，

△ APF is an isosceles right triangle,

So, the answer is: constant, isosceles right triangle;

(2)① solution: let AP=CQ=x, then BP=2-x,

∫S△PCQ =

1

four

S△ABC，

∴

1

2

x(2-x)= 1

1

four

×(

1

2

×2×2),

After sorting, x2-2x+ 1=0,

The solution is x= 1,

∴ap= 1；

A: The length of DE is constant and is a constant value.

Proof: As shown in the figure, QF⊥AC crosses the AC extension line at F,

Then < qcf = < ACB = < a = 45,

∵PE⊥AC，

∴∠AEP=90，

∴∠AEP=∠F=90，

The velocities of points p and q are equal,

∴AP=CQ，

At △QCF and △PAE,

∠QCF=∠A

AEP =∠F = 90°

Associated Press =CQ

∴△QCF≌△PAE(AAS)，

∴AE=CF,EP=QF，

∴AC=AE+EC=CF+EC=EF，

At △EPD and △FQD,

AEP =∠F = 90°

∠QDF =∠ Partial differential equation

EP=QF

∴△EPD≌△FQD(AAS)，

∴DE=DF，

∴DE=

1

2

EF=

1

2

Communication,

∫∠B = 90，AB=BC=2，

∴AC=

AB2+BC2

=

22+22

=2

2

∴DE=

1

2

×2

2

=

2

It is a fixed value.