As shown in Figure 1, it is known that P is a point on the diagonal AC of square ABCD (not coincident with a and C), PE ⊥ BC at point E, PF ⊥ CD at point F. (1) verification: BP = DP; (2) as shown in Figure 2, if quadrilateral PECF rotates counterclockwise around point C, is there always BP = DP in the process of rotation? If so, please prove it; if not, please explain it with counter examples; (3) try to select two vertices of square ABCD and connect them with the two vertices of quadrilateral PECF respectively, so that the length of the two segments is always equal when quadrilateral PECF rotates counterclockwise around point C, and prove your conclusion

As shown in Figure 1, it is known that P is a point on the diagonal AC of square ABCD (not coincident with a and C), PE ⊥ BC at point E, PF ⊥ CD at point F. (1) verification: BP = DP; (2) as shown in Figure 2, if quadrilateral PECF rotates counterclockwise around point C, is there always BP = DP in the process of rotation? If so, please prove it; if not, please explain it with counter examples; (3) try to select two vertices of square ABCD and connect them with the two vertices of quadrilateral PECF respectively, so that the length of the two segments is always equal when quadrilateral PECF rotates counterclockwise around point C, and prove your conclusion

(1) Proof: proof 1: in △ ABP and △ ADP, ∵ AB = ad ∠ BAC = ∠ DAC, AP = AP, ≌ ABP ≌ ADP, ≌ BP = DP. (2 points) proof 2: by using the square's axial symmetry, BP = DP. (2 points) (2 points) is not always true. (3 points) when the point P of quadrilateral PECF rotates to the BC side, DP > DC > BP, BP = DP is not true, when the point P of quadrilateral PECF is on the extension line of AC (3) connect be and DF, then be and DF are always equal. In Figure 1, it can be proved by square ABCD: AC bisection ∠ BCD, ∵ PE ⊥ BC, PF ⊥ CD, ≁ PE = PF, ≌ BCD = 90 °, the quadrilateral PECF is square. (7 points) ∵ CE = CF, ≂ DCF = ≂ BCE, BC = CD, ≌ BEC ≌ DFC, ≌ be = DF. (8 points)