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数学
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AB 是圆O的直径,P是弦AC的延长线上的一点,AC=PC,直线PB交圆O于点D,求CP=CD
人气:314 ℃ 时间:2019-08-19 14:17:11
解答
证明:AB为⊙O直径,∴BC⊥AP,又AC=PC
∴△ABP为等腰△
有∠A=∠P
∵A、C、B、D四点共圆
∴∠A=∠D
∴∠P=∠D
∴CP=CD
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