已知圆O1与圆O2外切于点A,AB是圆O1的直径,BD切与圆O2与点D,交圆O1与点c,求证AB×CD=AC×BD

问题描述:

已知圆O1与圆O2外切于点A,AB是圆O1的直径,BD切与圆O2与点D
,交圆O1与点c,求证AB×CD=AC×BD

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证:∵AB为直径∴∠ACB=90º又∠BDO₂=90º∴O₂D‖AC∴AB/AC=BO₂/O₂D又∵O₂D为小圆半径=AO₂故AB/AC=BO₂/O₂A=BD/CD故AB×CD=AC×BD