Al Biruni Problem -Solution

Let O be a circle of center A. C,B, and D are three points on the circle, B closer to D. Let F be the midpoint of the the arc CD. Let FG be a perpendicular to CB, G on CB. Then, CG = GB + BD
Solution: Let I be the intersection of FG with circle O(A). Let J be the intersection of ID with CB. Angle (BCI) ̂ is congruent to angle (BDI) ̂, since: CIDB is inscribed, and the opposite angles are supplementary, and angles (IDB) ̂ and (BDJ) ̂ are also supplementary. However, Since F is the midpoint of arc CD, then IG bisector of angle (CIJ) ̂ IG is perpendicular to CJ; therefore, IG is perpendicular bisector of CJ, and triangle CIJ is isosceles. Hence, angel (CIB) ̂ congruent to (IJC) ̂ Therefore: (IJC) ̂ = (BDI) ̂ = (DIB) ̂ So, triangle BDJ has BD = BJ Since CG = GJ and, GJ = GB + BJ= GB + BD then CG = GB +BD