点O是△ABC内的任意一点,作直线AO,BO,CO与边BC,CA,AB,分别交于点D,E,F
则BD/DC·CE/AE·
AF/BF=1.
证明:过A点作AN∥BE,AM∥CF分别交BC的延长线及反向延长线于点N、M.
因为AN∥BE,所以BD：BN=OD：AO,CE:AE=CB:BN·①
因为AM∥CF,所以CD:CM=OD:AO,AF:BF=CM:BC·②
因为BD:BN=OD:AO,CD:CM=OD:AO,所以CD:CM=BD:BN,即BD:CD=BN:CM.③
由①×②×③得:CE:AE×AF:BF×BD:CD=CB:BN×CM:BC×BN:CM=1,
即BD:DC·CE:AE·AF:BF=1