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

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.