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

The Length LM = L'M and the angle F is 90' in
both two triangle ABF and PFB' and the angle
B = B ' so these both tringle are equal .
Similarly other similar triangles are equal .
The length BF = B'F , CG = C'G KH = K'H .

M is the middle point of LL'. we joined CB'. DC'. ....,LK'

As in last proposition 21 we have also similar traingles in here ABF = FPB' and so the other are
Equal. Therefore,
BF / FA = B'F / FP = CG / PG = C'G / GQ = ...... = LM / RM;

After we sum up the equation we get

(BB' + CC'+................+LM) / AM = BF / FA =
A'B / BA