Let O be a circle of center A.[br]C,B, and D are three points on the circle, B closer to D.[br]Let F be the midpoint of the the arc CD.[br]Let FG be a perpendicular to CB, G on CB.[br]Then,[br]CG = GB + BD

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