已知:如图,⊙O和⊙ 内切于A,直线O 交⊙O 于另一点B、交⊙ 于另一点F,过B点作⊙ 的切线,切点为D

已知:如图,⊙O和⊙内切于A,直线O交⊙O于另一点B、交⊙于另一点F,过B点作⊙的切线,切点为D,交⊙O于C点,DE⊥AB,垂足为E.求证:(1)CD=DE;(2)若将两... 已知:如图,⊙O和⊙ 内切于A,直线O 交⊙O 于另一点B、交⊙ 于另一点F,过B点作⊙ 的切线,切点为D,交⊙O于C点,DE⊥AB,垂足为E.求证:(1)CD=DE;(2)若将两圆内切改为外切,其它条件不变,(1)中的结论是否成立?请证明你的结论.
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你的说明有点不清楚,不过图看得满清晰的。

我这样证明的,作辅助线,连接O1D,AD,AC。

因为 O1D垂直BC,AC垂直BC,

所以 O1D平行AC

所以 角O1DA=角DAC

又因为 三角形O1DA的两边O1D=O1A

所以 角O1AD(即角EAD)=角DAC

因此 三角形EDA全等于三角形CDA

因此 CD=DE

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如图解释,不是很难的

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连接AD,AC ,DF
AC垂直BC,DE垂直AF, <DFA=<ADC <ADF=90 所以<DAE=<DAC 且共斜边,故DE=DC
相等
AC垂直BC,DE垂直AF, <DFA=<ADC <ADF=90 所以<DAE=<DAC 且共斜边,故DE=DC
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