(1) It is easy to prove that DB=BC is an isosceles triangle and DB = BC, so the coordinates of point B (3,4) and point C (6,0) are obtained.

(2)DPB =∠A+∠ADP

∠DPB=∠DPQ + ∠BPQ =∠A + ∠BPQ

= = = = = & gt； ∠ADP = ∠BPQ

In addition, Democratic Action Party = PBQ

= = = = = & gt； ⊿ADP ∽ ⊿BPQ

= = = = = & gt； AP/AD = BQ/BP

= = = = = & gt； x/5 = (5-y)/(6-x)

= = = = = & gt； x^2 -6x -5y +25 =0

(3) Discuss in two situations.

① In the case of DP = DQ, according to the similarity of (2), there are:

AP/DP = BQ/PQ = = = & gt； x/y = (5-y)/[2y*cos∠A]

cos∠A=3/5

= = = = = & gt； 6x=25-5y

The simultaneous equation (2) is solved as follows: x = 0, y=5.

② In the case of DP = PQ, according to the similarity of (2), there are:

AP/DP = BQ/PQ = = = & gt； x=5-y

The equation of simultaneous (2) can be solved as follows: x = 1, y = 4.

(x=0 is also a solution, but at this time DP≠PQ does not meet the premise of this solution)

Based on the discussion of ① and ②, it is concluded that △PDQ is an isosceles triangle when x=0 or 1.