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[几何] 一道圆与边长的几何题

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tommywong Posted at 2014-3-25 17:04:20 |Read mode
捕获.PNG

如图,在⊙O中,弦AB与弦CD相交于G,OA⊥CD于点E,过B的直线与CD的延长线交于F,且AC//BF,
①若∠FGB=∠FBG,求证:BF为⊙O切线,
②若$tan∠F=\frac{3}{4},CD=8$,求半径,
③求证:$GF^2-GB^2=DF·GF$
现充已死,エロ当立。
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isee Posted at 2014-3-25 22:20:19
只看第三问,仅平行倒角即可,key:$GB^2=GD\cdot GF$;

事实上,F在CD直线上都成立,当然,得取绝对值。

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isee Posted at 2014-3-25 22:25:01
第2问,理论上,方法很多

一种笨笨方法供参考,还是平行倒一下角,
$Rt\triangle CEO$中,勾股得了,$(r-3)^2+4^2=r^2\Rightarrow r=\dfrac {25}6$

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isee Posted at 2014-3-25 22:31:41
难怪相对容易,2013年广东省茂名市中考数学第24题。

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