(2) Take the parallel lines whose intersection point P is AC and solve them according to the properties of the parallel lines;
(3) According to the different position of P, it is discussed in three situations.
Solution: Solution: (1) Solution 1: As shown in figure 1, extend the BP intersection AC to point E.
∵AC∥BD,∴∠PEA=∠PBD.
∠∠APB =∠PAE+∠PEA，
∴∠apb=∠pac+∠pbd；
Solution 2: As shown in Figure 2.
Passing point p is FP∑AC,
∴∠PAC=∠APF.
∵AC∥BD,∴FP∥BD.
∴∠FPB=∠PBD.
∴∠APB=∠APF+∠FPB
=∠PAC+∠PBD；
Solution 3: As shown in Figure 3,
∫AC∨BD，
∴∠CAB+∠ABD= 180，
∠PAC+∠PAB+∠PBA+∠PBD= 180。
∠ APB+∠ PBA+∠ PAB = 180，
∴∠APB=∠PAC+∠PBD.
(2) it is not established.
(3)(a)
When the moving point P is on the right side of ray BA, the conclusion is
∠PBD=∠PAC+∠APB。
(b) When the moving point p is on the ray BA,
The conclusion is ∠ PBD = ∠ PAC+∠ APB.
Or < PAC = < PBD+< APB or < APB = 0,
∠PAC=∠PBD (just write one)
(c) When the moving point p is to the left of the ray BA,
The conclusion is ∠ PAC = ∠ APB+∠ PBD.
Choose (a) to prove that:
As shown in fig. 4, connect PA, PB and AC to m.
∫AC∨BD，
∴∠PMC=∠PBD.
∠∠PMC =∠PAM+∠APM (one outer angle of a triangle is equal to the sum of two inner angles that are not adjacent),
∴∠PBD=∠PAC+∠APB.
Select (b) to prove: as shown in Figure 5.
∵ Point P is on Leiba, ∴∠APB=0 degrees.
∵AC∥BD,∴∠PBD=∠PAC.
∴∠PBD=∠PAC+∠APB
Or < PAC = < PBD+< APB.
Or ∠ APB = 0, ∠ PAC = ∠ PBD.
Option (c) proves that:
As shown in fig. 6, connect PA, PB and AC to F.
∵AC∥BD,∴∠PFA=∠PBD.
∠∠PAC =∠APF+∠PFA，
∴∠PAC=∠APB+∠PBD.
Comments: This question examines the nature of the angular bisector; It is an exploratory question, which aims to examine students' ability to analyze and study materials and master the properties of parallel lines and angular bisectors. Doing (1)(2) minor problems can provide ideas for (3) minor problems.