linearly a line parallel to that segment.  Thus, given that B is the middle point of AC, we may draw any two lines through A, and any line through B cutting them in points N and L.  Join N and L to C and get the points K and M on the two lines through A.  Then KM is parallel to AC. The bisection of a given segment and the drawing of a line parallel to the segment are equivalent data when linear construction is used.

42. It is not difficult to give a linear construction for the problem to divide a given segment into n equal parts, given only a parallel to the segment.  This is simple enough when n is a power of 2.  For any other number, such as 29, divide any segment on the line parallel to AC into 32 equal parts, by a repetition of the process just described.  Take 29 of these, and join the first to A and the last to C.  Let these joining lines meet in S.  Join S to all the other points.  Other problems, of a similar sort, are given at the end of the chapter.

43.  Numerical relations. Since three points, given in order, are sufficient to determine a fourth, as explained above, it ought to be possible to reproduce the process numerically in view of the one-to-one correspondence which exists between points on a line and numbers; a correspondence which, to be sure, we have not established here, but which is discussed in any treatise on the theory of point sets.  We proceed to discover what relation between four numbers corresponds to the harmonic relation between four points.

[Figure 8]

FIG. 8

44. 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’,

[formula]

the minus sign being given to the ratio on account of the fact 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.

45. Writing the last equation in the form

CB :  AB = -CD :  AD,

and using the fundamental relation connecting three points on a line,