Rt三角形ABC内接于圆O,AC=BC,∠BAC的平分线AD与圆O交于点D,与BC交于点E,延长BD,与AC的延长线交于点F,连结CD,G是CD中点,连结OG.若OG*DE=3(2-根号2),求圆O面积
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![Rt三角形ABC内接于圆O,AC=BC,∠BAC的平分线AD与圆O交于点D,与BC交于点E,延长BD,与AC的延长线交于点F,连结CD,G是CD中点,连结OG.若OG*DE=3(2-根号2),求圆O面积](/uploads/image/z/2638926-54-6.jpg?t=Rt%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E5%86%85%E6%8E%A5%E4%BA%8E%E5%9C%86O%2CAC%3DBC%2C%E2%88%A0BAC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BFAD%E4%B8%8E%E5%9C%86O%E4%BA%A4%E4%BA%8E%E7%82%B9D%2C%E4%B8%8EBC%E4%BA%A4%E4%BA%8E%E7%82%B9E%2C%E5%BB%B6%E9%95%BFBD%2C%E4%B8%8EAC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9F%2C%E8%BF%9E%E7%BB%93CD%2CG%E6%98%AFCD%E4%B8%AD%E7%82%B9%2C%E8%BF%9E%E7%BB%93OG.%E8%8B%A5OG%2ADE%3D3%EF%BC%882-%E6%A0%B9%E5%8F%B72%EF%BC%89%2C%E6%B1%82%E5%9C%86O%E9%9D%A2%E7%A7%AF)
Rt三角形ABC内接于圆O,AC=BC,∠BAC的平分线AD与圆O交于点D,与BC交于点E,延长BD,与AC的延长线交于点F,连结CD,G是CD中点,连结OG.若OG*DE=3(2-根号2),求圆O面积
Rt三角形ABC内接于圆O,AC=BC,∠BAC的平分线AD与圆O交于点D,与BC交于点E,延长BD,与AC的延长线交于点F,连结CD,G是CD中点,连结OG.
若OG*DE=3(2-根号2),求圆O面积
Rt三角形ABC内接于圆O,AC=BC,∠BAC的平分线AD与圆O交于点D,与BC交于点E,延长BD,与AC的延长线交于点F,连结CD,G是CD中点,连结OG.若OG*DE=3(2-根号2),求圆O面积
连接OD,交BC与点H,连接OC,
由CG=GD知OG⊥CD,故∠ODG=∠OCG=45°+22.5°=67.5°
又∠BED=∠EAB+∠EBA=67.5°
故∠ODG=∠BED,又∠OGD=∠BDE=90°
∴△OGD∽△BDE
得出OG/BD=GD/DE即BD*GD=OG*DE=3(2-sqrt2)
又GD=CD/2=BD/2,故BD^2=6(2-sqrt2)
又BD^2=BH^2+HD^2=BH^2+(OD-OH)^2
设AO=OB=OD=r,
则BH=OH=sqrt2r/2,
代入得(r^2/2)+[(2-sqrt2)r/2]^2=6(2-sqrt2)
解得r^2=6
故S=πr^2=6π.