In △ABC and △SBR, ∠ C = ∠BRS = 45, ∠ B is the common angle.

∴△ABC∽△SBR

(2) The length of line segment TS is equal to the length of PA.

∵ quadrilateral PTEF is a square,

∴pf=pt,∠spt+∠fpa= 180-∠TPF = 90，

In Rt△PFA, ∠PFA

+∠FPA=90，

∴∠PFA=∠TPS，

∴Rt△PAF≌Rt△TSP,∴PA=TS.

When the point p moves so that t and r coincide,

At this time, △PFA and △TSP are isosceles right triangles with equal bottoms, that is, PA = ts.

(3) From the meaning of the question, RS is the height on the base PB of the isosceles Rt△PRB.

∴PS=BS，

∴BS+PS+PA= 1，

∴PS=( 1-PA)/2

Let the length of PA be x, and it is easy to know that AF=PS.

Then y = pf 2 = pa 2+PS 2, y = x 2+(( 1-x)/2) 2,

That is y = 5/4x 2- 1/2x+ 1/4.

According to the property of quadratic function, when x = 1/5, y has the minimum value 1/5.

As shown in fig. 2, when the P point moves so that T and R coincide, PA = TS is the maximum.

Yi Zheng isosceles Rt△PAF≌ isosceles Rt△PSR≌ isosceles Rt△BSR,

∴PA= 1/3.

When p and a coincide, x = 0.

The value range of ∴x is 0≤x≤ 1/3.

① When the value of x increases from 0 to 1/5, the value of y decreases from 1/4 to 1/5.

② When the value of x increases from 1/5 to 1/3, the value of y increases from 1/5 to 2/9.

(Note: ① ② If you are right in any place, you will get 1 point, and if you are right in two places, you will get 1 point. )

∫ 1/5≤2/9≤ 1/4, ∴during the movement of point p.

The minimum value of the square PTEF area y is 1/5, and the maximum value of y is 1/4.