# 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

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