If a polygon be inscribed in a segment of a circle LAL' so that all its sides excluding the base are equal and their number even, as LK ... A ...K'L', A being the middle point of the segment; and if the lines BB', CC',... parallel to the base LL' and joining pairs of angular points be drawn, then (BB'+ CC'+..+ LM):AM = A'B:BA, where M is the middle point of LL' and AA' is the diameter through M [Fig. 6.6].
This is the geometric equivalent of the trigonometric equation
$$sin\frac {\theta}n +sin \frac{2\theta}n + ... sin\frac{n-1}n\theta + \frac12 sin\frac {n \theta}n= \frac{1-cos\theta}2cot\frac{\theta}{2n} $$
