every side of the triangle: and the two segments which terminated at any angular point of the triangle are never in the same product. Thus if wo begin one product with the segment BD, the other segment of the side BC, namely DC, occurs in the other product; then the segment CE occurs in the first product, so that the two segments CD and CE, which terminate at C, do not occur in the same product; and so on.
The student should for exercise draw another figure for the case in which the transversal meets all the sides produced, and obtain the same result.
58. Conversely, it may be shewn by an indirect proof that if the product BD.CE.AF be equal to the product DC.EA.FB, the three points D,E,F lie in the same straight line.
59. If three straight lines he drawn through the angular points of a triangle to the opposite sides, and meet at the same pointy the product of three segments in order is equal to the product of the other three segments.
Let ABC be a triangle. From the angular points to the opposite sides let the straight lines AOD, BOE, COF be drawn, which meet at the point O: the product AF.BD.CE shall be equal to the product FB.DC.EA.
For the triangle ABD is cut by the transversal FOC, and therefore by the theorem in 57 the following products are equal, AF.'BC.DO, and FB.CD.OA.
Again, the triangle ACD is cut by the transversal EOB, and therefore by the theorem in 57 the following products are equal, AO.DB.CE and OD.BC.EA.
