A standard proof A proof given by John Wellesley Russell uses
Pasch's axiom to consider cases where a line does or does not meet a triangle. First, the sign of the
left-hand side will be negative since either all three of the ratios are negative, the case where the line misses the triangle (see diagram), or one is negative and the other two are positive, the case where crosses two sides of the triangle. To check the magnitude, construct perpendiculars from to the line and let their lengths be respectively. Then by
similar triangles it follows that \left|\frac{\overline{AF}}{\overline{FB}}\right| = \left|\frac{a}{b}\right|, \quad \left|\frac{\overline{BD}}{\overline{DC}}\right| = \left|\frac{b}{c}\right|, \quad \left|\frac{\overline{CE}}{\overline{EA}}\right| = \left|\frac{c}{a}\right|. Therefore, \left|\frac{\overline{AF}}{\overline{FB}}\right| \times \left|\frac{\overline{BD}}{\overline{DC}}\right| \times \left|\frac{\overline{CE}}{\overline{EA}}\right| = \left| \frac{a}{b} \times \frac{b}{c} \times \frac{c}{a} \right| = 1. For a simpler, if less symmetrical way to check the magnitude, draw parallel to where meets at . Then by similar triangles \left|\frac{\overline{BD}}{\overline{DC}}\right| = \left|\frac{\overline{BF}}{\overline{CK}}\right|, \quad \left|\frac{\overline{AE}}{\overline{EC}}\right| = \left|\frac{\overline{AF}}{\overline{CK}}\right|, and the result follows by eliminating from these equations. The converse follows as a corollary. Let be given on the lines so that the equation holds. Let be the point where crosses . Then by the theorem, the equation also holds for . Comparing the two, \frac{\overline{AF}}{\overline{FB}} = \frac{\overline{AF'}}{\overline{F'B}}\ . But at most one point can cut a segment in a given ratio so
A proof using homotheties The following proof uses only notions of
affine geometry, notably
homotheties. Whether or not are collinear, there are three homotheties with centers that respectively send to , to , and to . The composition of the three then is an element of the group of homothety-translations that fixes , so it is a homothety with center , possibly with ratio 1 (in which case it is the identity). This composition fixes the line if and only if is collinear with (since the first two homotheties certainly fix , and the third does so only if lies on ). Therefore are collinear if and only if this composition is the identity, which means that the magnitude of the product of the three ratios is 1: \frac{\overrightarrow{DC}}{\overrightarrow{DB}} \times \frac{\overrightarrow{EA}}{\overrightarrow{EC}} \times \frac{\overrightarrow{FB}}{\overrightarrow{FA}} = 1, which is equivalent to the given equation. ==History==