如图,已知a,b,c三点在半径为2的圆o上,ob与ac相交于d,若∠acb=∠aoc,则(1/bd)-(1/bc)=
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如图,已知a,b,c三点在半径为2的圆o上,ob与ac相交于d,若∠acb=∠aoc,则(1/bd)-(1/bc)=
![](http://img.wesiedu.com/upload/f/1f/f1f15a47e5cabb3a80e97739883199df.jpg)
![](http://img.wesiedu.com/upload/f/1f/f1f15a47e5cabb3a80e97739883199df.jpg)
![如图,已知a,b,c三点在半径为2的圆o上,ob与ac相交于d,若∠acb=∠aoc,则(1/bd)-(1/bc)=](/uploads/image/z/5287683-3-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E4%B8%89%E7%82%B9%E5%9C%A8%E5%8D%8A%E5%BE%84%E4%B8%BA2%E7%9A%84%E5%9C%86o%E4%B8%8A%2Cob%E4%B8%8Eac%E7%9B%B8%E4%BA%A4%E4%BA%8Ed%2C%E8%8B%A5%E2%88%A0acb%3D%E2%88%A0aoc%2C%E5%88%99%281%2Fbd%29-%281%2Fbc%29%3D)
∵∠acb=∠oac
∴ oa//bc
△AOD∽△CBD
bc/bd=oa/od
bc/bd-1=oa/od-1
(bc-bd)/bd=(oa-od)/od
=bd/od
=bc/oa
(bc-bd)/(bdxbc)=1/oa
=1/2
(1/bd)-(1/bc)=(bc-bd)/(bdxbc)
=1/2
再问: 请问(bc-bd)/(bdxbc)=1/oa这一步是怎么来的呢
再答: 从上面推出: (bc-bd)/bd=bc/oa (bc-bd)/(bdxbc)=1/oa
∴ oa//bc
△AOD∽△CBD
bc/bd=oa/od
bc/bd-1=oa/od-1
(bc-bd)/bd=(oa-od)/od
=bd/od
=bc/oa
(bc-bd)/(bdxbc)=1/oa
=1/2
(1/bd)-(1/bc)=(bc-bd)/(bdxbc)
=1/2
再问: 请问(bc-bd)/(bdxbc)=1/oa这一步是怎么来的呢
再答: 从上面推出: (bc-bd)/bd=bc/oa (bc-bd)/(bdxbc)=1/oa
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