By Derrick Norman Lehmer

Meant to offer, As easily As attainable, The necessities of man-made Projective Geometry - Chapters: One-To-One Correspondence - family among primary kinds In One-To-One Correspondence With one another - mix of 2 Projectively comparable primary varieties - Point-Rows Of the second one Order - Pencils Of Rays Of the second one Order - Poles And Polars - Metrical homes Of The Conic Sections - Involution - Metrical homes Of Involutions - at the background of man-made Projective Geometry - Index

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**Additional info for An Elementary Course In Synthetic Projective Geometry **

**Example text**

Let A and A', B and B', be corresponding points, and let also the point M of intersection of u and u' correspond to itself. Let AA' and BB' meet in the point S. Take S as the center of two pencils, one perspective to u and the other perspective to u'. In these two pencils SA coincides with its corresponding ray SA', SB with its corresponding ray SB', and SM with its corresponding ray SM'. The two pencils are thus identical, by the preceding theorem, and any ray SD must coincide with its corresponding ray SD'.

Four harmonic planes. We also define four harmonic planes as four planes through a line which pass one through each of four harmonic points, and we may show that Four harmonic planes are cut by any plane not passing through their common line in four harmonic lines, and also by any line in four harmonic points. For let the planes ±, ², ³, ´, which all pass through the line g, pass also through the four harmonic points A, B, C, D, so that ± passes through A, etc. Then it is clear that any plane À through A, B, C, D will cut out four harmonic lines from the four planes, for they are lines through the intersection P of g with the plane À, and they pass through the given harmonic points A, B, C, D.

We shall [22] [23] 24 An Elementary Course in Synthetic Projective Geometry therefore avail ourselves of one that lends itself most readily to the solution of the problem. We choose the point L so that the triangle ALC is isosceles (Fig. 7). 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.