BC = BD + DC, CA = CD + DA, AB = AD + DB;
or multiplying these by AD, BD, CD respectively, we get
| BC · AD = BD · AD + DC · AD = BD · AD − CD · AD CA · BD = CD · BD + DA · BD = CD · BD − AD · BD AB · CD = AD · CD + DB · CD = AD · CD − BD · CD. |
It will be seen that the sum of the right-hand sides vanishes, hence that
BC · AD + CA · BD + AB · CD = 0
(3)
for any four points on a line.
| Fig. 3. |
§ 11. If C is any point in the line AB, then we say that C divides the segment AB in the ratio AC/CB, account being taken of the sense of the two segments AC and CB. If C lies between A and B the ratio is positive, as AC and CB have the same sense. But if C lies without the segment AB, i.e. if C divides AB externally, then the ratio is negative. To see how the value of this ratio changes with C, we will move C along the whole line (fig. 3), whilst A and B remain fixed. If C lies at the point A, then AC = 0, hence the ratio AC : CB vanishes. As C moves towards B, AC increases and CB decreases, so that our ratio increases. At the middle point M of AB it assumes the value +1, and then increases till it reaches an infinitely large value, when C arrives at B. On passing beyond B the ratio becomes negative. If C is at P we have AC = AP = AB + BP, hence
| AC | = | AB | + | BP | = − | AB | − 1. |
| CB | PB | PB | BP |