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.

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

由题义知只需证明角CAF=角BAO,或者角FAD=角OAE既可