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如图,平行四边形ABCD中,AB=5,AD=3,AE平分∠DAB交BC的延长线于...
如图,平行四边形ABCD中,AB=5,AD=3,AE平分∠DAB交BC的延长线于F点,则CF=
.
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131101193314600178660/SYS201311011933146001786011_ST/images0.png)
考点分析:
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已知:如图,在直角坐标系中,有菱形OABC,A点的坐标为(10,0),对角线OB、AC相交于D点,双曲线
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(x>0)经过D点,交BC的延长线于E点,且OB•AC=160,有下列四个结论:
①双曲线的解析式为
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131101193314600178660/SYS201311011933146001786009_ST/1.png)
(x>0);
②E点的坐标是(4,8);
③sin∠COA=
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;
④AC+OB=
![manfen5.com 满分网](http://img.manfen5.com/res/CZSX/web/STSource/20131101193314600178660/SYS201311011933146001786009_ST/3.png)
,其中正确的结论有( )
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