To begin the proof, perpendiculars are constructed from the vertices A, B, and C to the line through D, E, and F (Figure 9). Consider three pairs of similar triangles: AHaF and BHbF, CHcD and BHbD, and AHaE and CHcE. The lengths of the constructed perpendicular segments are x, y, and z. Now, from the similar triangles, we have AF/BF= x/z, BD/CD= z/y, and CE/AE= y/x. Multiplying these we have AF/BF*BD/CD*CE/AE=(xzy)/(zyx)=1. Also, AF*BD*CE= BF*CD*AE which is the alternate statement of Menelaus' Theorem.