Clearing of fractions by multiplying throughout by AB · BC · CA, we have to prove
−AD · AX · BC − BD · BX · CA − CD · CX · AB = AB · BC · CA.
Take A as origin and let AB = a, AC = b, AD = c, AX = x. Substituting for the segments in terms of a, b, c, x, we obtain on simplification
a²b − ab² = −ab² + a²b, an obvious identity.
An alternative method of testing a relation is illustrated in the following example:— If A, B, C, D, E, F be six collinear points, then
| AE · AF | + | BE · BF | + | CE · CF | + | DE · DF | = 0. |
| AB · AC · AD | BC · BD · BA | CD · CA · CB | DA · DB · DC |
Clearing of fractions by multiplying throughout by AB · BC · CD · DA, and reducing to a common origin O (calling OA = a, OB = b, &c.), an equation containing the second and lower powers of OA ( = a), &c., is obtained. Calling OA = x, it is found that x = b, x = c, x = d are solutions. Hence the quadratic has three roots; consequently it is an identity.
The relations connecting five points which we have instanced above may be readily deduced from the six-point relation; the first by taking D at infinity, and the second by taking F at infinity, and then making the obvious permutations of the points.]
Projection and Cross-ratios
§ 12. If we join a point A to a point S, then the point where the line SA cuts a fixed plane π is called the projection of A on the plane π from S as centre of projection. If we have two planes π and π′ and a point S, we may project every point A in π to the other plane. If A′ is the projection of A, then A is also the projection of A′, so that the relations are reciprocal. To every figure in π we get as its projection a corresponding figure in π′.