A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

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.
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.
(2)
280 TRIGONOMETRY
The equation in fact gives the formula,
[crd. («-£)}. (crd. 180°) = (crd. a) .
which is, of course, equivalent to
[crd. (180°-j8)}
-(crd. /3).{crd. (180°-a)},
sin ($ — 0) = sin cos — cos sin 0, where a — 2 0, (3 = 20.
By means of this formula Ptolemy obtained
crd. 12° = crd. (72°-60°) = 12P 32' 36".
(8) Equivalent of sin 2 J0 = § (1— cos0).
But, in order to get the chords of smaller angles still, we
want a formula for finding the chord of half an arc when the
chord of the arc is given. This is the subject of Ptolemy's
next proposition.
Let BG be an arc of a circle with diameter AC, and let the
arc BC be bisected at D. Given (crd. BG), it is required to
find (crd. DC).
Draw DF perpendicular to AG,
and join AB, AD, BD, DC. Measure
AE along AG equal to AB, and join
BE.
Then shall FG be equal to EF, or
FG shall be half the difference between
AC and AB.
For the triangles A BD, AED are
equal in all
respects, since two sides
of the one are equal to two sides of the other and the included
angles BAD, EAD, standing on equal arcs, are equal.
Therefore
ED = BD = DC,
and the right-angled triangles DEF, DCF are equal in all
respects, whence EF = FG, or CF = ±(AC-AB).
Now
AC.CF=CD\
whence (crd. CDf = \AC (AG-AB)
= i(crd. 180 o ).{(crd.l80 o )-(erd.l80 o -£(7)}.
This is, of course, equivalent to the formula
sin 2 i