如图 连接AO并延长,于DC交于点R,于BC的延长线交于点S,若AD=4,∠DCB=60°,BS=10求AS和OR
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如图 连接AO并延长,于DC交于点R,于BC的延长线交于点S,若AD=4,∠DCB=60°,BS=10求AS和OR
四边形ABCD是菱形
![](http://img.wesiedu.com/upload/d/fd/dfdddf80eb5278b10fa880d157473f0c.jpg)
四边形ABCD是菱形
![](http://img.wesiedu.com/upload/d/fd/dfdddf80eb5278b10fa880d157473f0c.jpg)
![如图 连接AO并延长,于DC交于点R,于BC的延长线交于点S,若AD=4,∠DCB=60°,BS=10求AS和OR](/uploads/image/z/19365121-1-1.jpg?t=%E5%A6%82%E5%9B%BE+%E8%BF%9E%E6%8E%A5AO%E5%B9%B6%E5%BB%B6%E9%95%BF%2C%E4%BA%8EDC%E4%BA%A4%E4%BA%8E%E7%82%B9R%2C%E4%BA%8EBC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9S%2C%E8%8B%A5AD%EF%BC%9D4%2C%E2%88%A0DCB%EF%BC%9D60%C2%B0%2CBS%EF%BC%9D10%E6%B1%82AS%E5%92%8COR)
你要的答案是:
(1)由ABCD是菱形和∠BCD=60°可知∠ABCA=120°且AB=BC=CD=DA=4
过点A做BC的垂线交于H ∴∠ABH=60° 又∵AB=4 所以可知在Rt△ABH中BH=2 AH=2√3 ∴SH=12 由勾股定理知AS=2√39
易证:△ADO∽△SBO 所以AD/SB=OD/OB=AO/SO ④ 又∵△ABO∽△RDO ∴OD/OB=OR/OA ⑤
由④⑤得OR/OA=AD/SB=4/10⑥ 又∵AO/SO=4/10⑦且 AO+SO=AS=2√39⑧
由⑥⑦⑧得OR=4/35AS=8/35√39
(1)由ABCD是菱形和∠BCD=60°可知∠ABCA=120°且AB=BC=CD=DA=4
过点A做BC的垂线交于H ∴∠ABH=60° 又∵AB=4 所以可知在Rt△ABH中BH=2 AH=2√3 ∴SH=12 由勾股定理知AS=2√39
易证:△ADO∽△SBO 所以AD/SB=OD/OB=AO/SO ④ 又∵△ABO∽△RDO ∴OD/OB=OR/OA ⑤
由④⑤得OR/OA=AD/SB=4/10⑥ 又∵AO/SO=4/10⑦且 AO+SO=AS=2√39⑧
由⑥⑦⑧得OR=4/35AS=8/35√39