Ceva's Theorem | Geometry | Math Olympiad | I.S.I. Entrance

Given a triangle ABC, let the lines AP, BP and CP be drawn from the vertices to a common point P (not on one of the sides of ABC), to meet opposite sides at D, E and F respectively. (The segments AD, BE, and CF are known as cevians.) Then, using signed lengths of segments, AF/FB * BD/DC * CE/EA =1. Problem useful for I.S.I B.Stat B.Math Entrance, CMI Entrance and Math Olympiad