如图,在平行四边形ABCD中,AC,DB为对角线,作AE⊥CB于E,AF⊥CD于F,延长FE,DB交于O,连结AO.
2个回答

梅捏老师定理

(CE/EB)*(BO/OD)*(DF/FC)=1

转化面积比

(S△CAE/S△BAE)(S△BAO/S△OAD)(S△DAF/S△FAC)=1

利用面积公式

2S△CAE=AC*AE*sin角CAE

转化三角比

(sin角CAE/sin角BAE)(sin角BAO/sin角OAD)(sin角DAF/sin角FAC)=1

设

角ACE=x

角ACF＝y

角OAD＝?

[sin(90-y)/sin(x+y-90)][sin(x+y+?)/sin(?)][sin(x+y-90)/sin(90-x)]=1

[sin(90-y)/sin(90-x)][sin(x+y+?)/sin(?)]=1

sin(90-y)sin(x+y+?)=sin(?)sin(90-x)

?＝90－y

所以AC⊥AO

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