A history of Greek mathematics - Wilbourhall.org

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PTOLEMY'S SYNTAXIS 279
The proposition giving the required formula depends upon
a lemma, which is the famous Ptolemy's theorem '
'.
Given a quadrilateral ABCD inscribed in a circle, the
diagonals being AG, BD, to prove that
AC.BD = AB.DC+AD.BC.
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 A BE, DBG are
equiangular, and therefore
AB:AE=BD:DC,
or AB.DC= 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 EBC, and the
triangles ABD, EBC are equiangular
therefore BC :
or
CE = BD: DA,
AD.BC = CE.BD.
(2)
By adding (1)
and (2), we obtain
AB.DC+AD.BC=AC. BD.
Now let AB, AC be two arcs terminating at A, the extremity
of the diameter. AD of a circle, and let
AG (= oc) be greater than AB (=/?;.
Suppose that (crd. AC) and (crd. AB)
are given : it is required to find
(crd. BC).
Join BD y CD.
Then, by the above theorem,
AC .BD = BC .
AD + AB .CD.
Now AB, AC are given"; therefore BD = crd. (180° -AB)
and CD = crd. (180° — A C) are known. And AD is known.
Hence the remaining chord BC (crd. BC) is known.