(2) As shown in the figure, extend DQ to point E, make EQ=DQ, and connect AE and BE.

Because point Q is the midpoint of AC, ∠AQE=∠CQD, EQ=DQ,

So △ AQE △ CQD (SAS), CD=AE, ∠DCQ=∠EAQ,

And because the problem (1) has proved that △PAB∽△PCD has AB/CD=BP/DP, AB/AE = BP/DP 1,

△BEQ has BAE = 360-BAQ-EAQ = 360-BAQ-DCQ.

There are ∠ BPD = 540-∠ BAQ-∠ DCQ-∠ ABP-∠ CDP = 360-∠ BAQ-∠ DCQ in the pentagonal ABPDC.

So ∠BAE=∠BPD②, then △BAE∽△BPD can be known from ① ②, and ∠ABE=∠PBD.

So ∠ EBD = ∠ ABP = 90, because EQ = DQ△EBD in the right angle, that is, BQ is the center line on the hypotenuse.

So we can know that BQ=DQ=EQ from "the midline on the hypotenuse of a right triangle is equal to half of the hypotenuse".

(3), because the problem (2) has proved that △BAE∽△BPD has AB/PB=BE/BD and ∠ EBD = ∠ ABP = 90.

So △EBD∽△ABP has △ EDB = △ APB, because BQ=DQ, that is, △BQD is an isosceles triangle.

So ∠ bqd =180-2 ∠ EDB = 2× (90-∠ EDB) = 2× (90-∠ APB) = 2 ∠ PAB.