By Derrick Norman Lehmer
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Extra resources for An Elementary Course in Synthetic Projective Geometry
Since D is supposed to be at infinity, the line KM is parallel to AC. Therefore the triangles KAC and MAC are equal, and the triangle ANC is also isosceles. The triangles CNL and ANL are therefore equal, and the line LB bisects the angle ALC. B is therefore the middle point of AC, and we have the theorem The harmonic conjugate of the middle point of AC is at infinity. FIG. 7 41. Parallels and mid-points 25 40. Projective theorems and metrical theorems. Linear construction. This theorem is the connecting link between the general protective theorems which we have been considering so far and the metrical theorems of ordinary geometry.
Correspondence between harmonic conjugates. Given four harmonic points, A, B, C, D; if we fix A and C, then B and 38. Separation of harmonic conjugates 23 D vary together in a way that should be thoroughly understood. To get a clear conception of their relative motion we may fix the points L and M of the quadrangle K, L, M, N (Fig. 6). Then, as B describes the point-row AC, the point N describes the point-row AM perspective to it. Projecting N again from C, we get a pointrow K on AL perspective to the point-row N and thus projective to the point-row B.
Let A, B, C, D be four harmonic points (Fig. 8), and let SA, SB, SC, SD be four harmonic lines. Assume a line drawn through B parallel to SD, meeting SA in A' and SC in C'. Then A', B', C', and the infinitely distant point on A'C' are four harmonic points, and therefore B is the middle point of the segment A'C'. Then, since the triangle DAS is similar to the triangle BAA', we may write the proportion AB : AD = BA' : SD. Also, from the similar triangles DSC and BCC', we have CD : CB = SD : B'C. From these two proportions we have, remembering that BA' = BC', AB · CD = −1, AD · CB the minus sign being given to the ratio on account of the fact  28 An Elementary Course in Synthetic Projective Geometry that A and C are always separated from B and D, so that one or three of the segments AB, CD, AD, CB must be negative.