摘要:∵ 点P在射线BA上.∴∠APB = 0°.∵ AC∥BD , ∴∠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 ,

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