A history of Greek mathematics - Wilbourhall.org

322 HERON OF ALEXANDRIA
chap. 30 of the Dioptra ;
but it is now known (from Arabian
sources) that the proposition is due to Archimedes.
Let the sides of the triangle ABO be given in length.
Inscribe the circle DEF, and let
Join AO, BO, GO, DO, EO, FO.
Then
BC.0D=2AB0C,
GA.0E = 2AC0A,
AB.0F=2AA0B:
be the centre.
whence, by addition,
p.0D = 2 A ABO,
where p
is the perimeter.
Produce CB to H, so that BH = AF.
Then, since AE = AF, BF = BD, and CE = OD, we have
CH=±p = s.
Therefore
CH.0D = AABC.
But CH.OD is the 'side' of the product GH 2 .OD 2 , i.e.
V(CH 2 . OD 2 ),
so that (AABC) 2 = CH 2 .0D 2 .