If a straight line intersects with three sides AB, BC, CA of △ABC or its extension lines at points F, D and E, then (AF/FB )× (BD/DC )× (CE/EA) =1. Or: let x, y and z be on the straight lines of BC, CA and AB of △ABC, then the necessary and sufficient conditions of x, y and Z*** lines are (az/zb) * (bx/xc) * (cy/ya) =1.

Certificate: ABC draws a vertical line AA'BB'CC' to three sides after three o'clock, so AD: DB = AA': BB', BE: EC = BB': CC', CF: FA = CC': AA' SO (AF/FB) × (BD/DC )× (CE/EA) =