如图,直线AC平行BD,连接AB,BD及线段AB把平面分成1,2,3,4四个部分,规定：线上各点不属于任何部分,当动点P

（ 1）解法一：如图 9- 1 延长BP交 直线 AC 于点E ∵ AC ∥ BD ,∴∠PEA = ∠PBD .∵∠APB = ∠PAE + ∠PEA ,∴∠APB = ∠PAC + ∠PBD .解法二：如图 9- 2 过点P作FP∥ AC ,∴∠PAC = ∠APF .∵ AC ∥ BD ,∴FP∥ BD .∴∠FPB =∠PBD .∴∠APB =∠APF +∠FPB =∠PAC + ∠PBD .解法三：如图 9- 3 ,∵ AC ∥ B D ,∴∠CAB +∠ABD = 180° 即∠PAC +∠PAB +∠PBA +∠PBD = 180°.又∠APB +∠PBA +∠PAB = 180°,∴∠APB =∠PAC +∠PBD .（ 2 ）不成立.（ 3 ）(a)当动点P在射线BA的右侧时,结论是 ∠PBD=∠PAC+∠APB .(b)当动点P在射线BA上,结论是∠PBD =∠PAC +∠APB .或∠PAC =∠PBD +∠APB 或∠APB = 0°,∠PAC =∠PBD（任写一个即可）.(c) 当动点P在射线BA的左侧时,结论是∠PAC =∠APB +∠PBD .选择(a) 证明：如图 9- 4 ,连接PA,连接PB交 AC 于M ∵ AC ∥ B D ,∴∠PMC =∠PBD .又∵∠PMC =∠PAM +∠APM ,∴∠PBD =∠PAC +∠APB .选择(b) 证明：如图 9-5 ∵点P在射线BA上,∴∠APB = 0°.∵ AC ∥ B D ,∴∠PBD =∠PAC .∴∠PBD =∠PAC +∠APB 或∠PAC =∠PBD+∠APB 或∠APB = 0°,∠PAC =∠PBD.选择(c) 证明：如图 9-6,连接PA,连接PB交 AC 于F ∵ AC ∥ BD ,∴∠PFA =∠PBD .∵∠PAC =∠APF +∠PFA ,∴∠PAC =∠APB +∠PBD .