Proof.—Let a, b, c denote the lines AA′, BB′, CC′, which meet at S. Then these may be taken as bases of projective rows, so that A, A′, S on a correspond to B, B′, S on b, and to C, C′, S on c. As the point S is common to all, any two of these rows will be perspective.
| If | S1 be the centre of projection of rows | b and c, |
| S2 ” ” ” | c and a, | |
| S3 ” ” ” | a and b, |
and if the line S1S2 cuts a in A1, and b in B1, and c in C1, then A1, B1 will be corresponding points in a and b, both corresponding to C1 in c. But a and b are perspective, therefore the line A1B1, that is S1S2, joining corresponding points must pass through the centre of projection S3 of a and b. In other words, S1, S2, S3 lie in a line. This is Desargues’ celebrated theorem if we state it thus:—
Theorem of Desargues.—If each of two triangles has one vertex on each of three concurrent lines, then the intersections of corresponding sides lie in a line, those sides being called corresponding which are opposite to vertices on the same line.
The converse theorem holds also, viz.
Theorem.—If the sides of one triangle meet those of another in three points which lie in a line, then the vertices lie on three lines which meet in a point.
The proof is almost the same as before.
§ 38. Metrical Relations between Projective Rows.—Every row contains one point which is distinguished from all others, viz. the point at infinity. In two projective rows, to the point I at infinity in one corresponds a point I′ in the other, and to the point J′ at infinity in the second corresponds a point J in the first. The points I′ and J are in general finite. If now A and B are any two points in the one, A′, B′ the corresponding points in the other row, then
(AB, JI) = (A′B′, J′I′),
or