∵ AC∥BD, ∴ ∠PEA = ∠PBD.
∴ ∠APB = ∠PAC + ∠PBD.
解法三:如图3,
∵ AC∥BD,∴ ∠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4,连接PA,连接PB交AC于M
∵ AC∥BD,
∴ ∠PMC =∠PBD.
又∵∠PMC =∠PAM +∠APM,
∴ ∠PBD =∠PAC +∠APB.