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.