[UPDATE: I have made a translation and commentary of a German translation of this book. It is here. ]
Here is the general setup:
Let A and C be any two points on a circle. Let M be equidistant from both so that arc AM is equal to arc MC. Let B be another point on the circle, lying between A and M, so that AB < BC.Drop the perpendicular from M to F on BC.
We claim that AB + BF = FC.
I will not spoil the fun by giving the proofs here. But these are eight constructions I know about.
Draw E such that AB = EC. (As an alternative approach, draw E such that BF = FE).
Draw the rectangle such that H is on the circle. Extend BC such that DF = FC. Draw E such that BF = FE and D such that BM = MD. Draw E such that BF = FE and extend ME to G. Extend BC and MF and draw DAG colinear. At this point, I discovered a German translation of al Biruni's book (by Suter, link below). Therefore, I switched notation to match his figures. I can select the text in Preview, then Google Translate does a good job with it.Extend BG as shown.
Extend the perpendicular DE as shown. Draw AG. Draw the diameter DK. (Hint: DK is perpendicular to AG). Sources: Drakaki, al Biruni, Suter. There is a chapter in my geometry book on this. The chapter as a pdf is here, and the github repo for the book is here.