摘要:∴ ∠FPB =∠PBD . ∴ ∠APB =∠APF +∠FPB =∠PAC + ∠PBD .解法三:如图9-3.∵ AC∥BD , ∴ ∠CAB +∠ABD = 180° 即 ∠PAC +∠PAB +∠PBA +∠PBD = 180°. 又∠APB +∠PBA +∠PAB = 180°, ∴ ∠APB =∠PAC +∠PBD . (2)不成立.

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