(1) as the auxiliary line AC, the angle B = 60 degrees, AB=AC, and the triangle AB = AC is an equilateral triangle.

Angle b = angle ACD = 60 degrees.

AB=AC

Angle BAC= angle BAP = 60, then angle BAP = 60°- angle PAC= = angle CAQ.

The congruence of triangle ABP and ACQ can be obtained.

Therefore, if AP = AQ and the angle PAQ=60 degrees, the triangle APQ is an equilateral triangle.

(2) When crossing point A, AE is perpendicular to BC and E. In triangle APE, y 2 = (2-x) 2+12 is obtained by pythagorean theorem.

0 & lt= x & lt=4

(3) If PD⊥AQ, then PD is the median vertical line of △APQ (three lines are equal and three lines are one).

That is, PD bisects AQ vertically and the intersection point is O.

Then in △ADQ, DO is the vertical line of AQ, then dq = da = 4.

Then q coincides with c or CQ = 2cd = 8.

(1) If q and c overlap, then b and p overlap, and BP = 0.

② If Q is on the extension line of the line segment CD, it is at △ADQ, AD = DQ = 4, and the angle D = 120, so it is easy to get AQ = 4 root number 3, AP = AQ = 4 root number 3, that is, Y = 4 root number 3.

Substitute into the equation in 2) and discard the negative value to get x = 8.

In summary, the value of BP is 0 or 8.