The straight line AC//BD connecting AB, BD and AB divides the plane into four parts: 1, 2, 3, 4. It is stipulated that every point on a straight line does not belong to any part. When the moving point P falls on a certain part, it connects PA and PB to form angle PAC, angle APB and angle PBD. (Hint: The angle between two coincident rays and a common point is 0. )

(1) When the moving point p falls in the part of 1, it is proved that: angle APB= angle PAC+ angle PBD；;

(2) Whether the angle APB= angle PAC+ angle PBD holds when the moving point P is divided into two;

(3) When the moving point P is divided into three parts, fully explore the relationship between angle APB= angle PAC+ angle PBD, write the specific position of the moving point P and the corresponding conclusion, and choose a conclusion to prove it. Please draw your own picture.

(1) Proof: 1 Part lies between two parallel lines. Pass p for PQ∑AC, pass AB for q,

Then ∠APQ=∠PAC, ∠BPQ=∠PBD,

∴∠APB=∠APQ+∠BPQ=∠PAC+∠PBD.

(2) The second part is also between two parallel lines, and ∴∠APB=∠PAC+∠PBD holds.

(3) The third part is on the outer side of two parallel lines near AC. If PB passes through AC to E, then

∠PBD=∠PEC=∠PAC+∠APB or 180-(∠ PAC+∠ APB).