PTOLEMY’S SYNTAX IS 
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The proposition giving the required formula depends upon 
a lemma, which is the famous ‘ Ptolemy’s theorem 
Given a quadrilateral A BCD inscribed in a circle, the 
diagonals being AC, BD, to prove that 
AC. BD = AB.DG+AD. BG. 
The proof is well known. Draw BE so that the angle ABE 
is equal to the angle DBG, and let BE 
meet AG in E. 
Then the triangles ABE, DBG are 
equiangular, and therefore 
AB : AE = BD: DC, 
or AB.DG = AE.BD. (1) 
Again, to each of the equal angles 
ABE, DBG add the angle EBD; 
then the angle ABD is equal to the angle EBG, and the 
triangles ABD, EBG are equiangular; 
therefore BG: GE = BD : DA, 
or AD .BG = GE. BD. (2) 
By adding (1) and (2), we obtain 
AB.DG+AD.BG = AG.BD. 
Now let AB, AG be two arcs terminating at A, the extremity 
of the diameter AD of a circle, and let 
AG (= a) be greater than AB (=/?;. 
Suppose that (crd. AG) and (crd. AB) 
are given: it is required to find 
(crd. BO). 
Join BD, CD. 
Then, by the above theorem, 
AG.BD = BG.AD + AB.CD. 
Now AB, AG are given; therefore BD = crd. (180°— AB) 
and CD = crd. (180°-AG) are known.. And AD is known. 
Hence the remaining chord BG (crd. BG) is known.