PROOF OF THE FORMULA OF HERON’ 323 
Draw OL at right angles to OC cutting BC in K, and BL at 
right angles to BG meeting OL in L. Join CL. 
Then, since each of the angles COL, CBL is right, COBL is 
a quadrilateral in a circle. 
Therefore ¿COB + ¿CLB = 2 B. 
But ¿COB + ZAOF = 2R, because AO, BO, CO bisect the 
angles round 0, and the angles COB, AOF are together equal 
to the angles AOC, BOF, while the sum of all four angles 
is equal to 4 R. 
Consequently Z AOF = Z CLB. 
Therefore the right-angled triangles AOF, CLB are similar ; 
therefore BG: BL = AF: FO 
= BH: OD, 
and, alternately, CB : BH = BL : OD 
=BK:KD; 
whence, componendo, CH: HB = BD: DK. 
It follows that 
CH*: CH. HB = BD . DC: CD . DK 
= BD. DC: OD 2 , since the angle COK is right. 
Therefore (A ABC) 2 — CLP . OD 2 (from above) 
= CH .HB .BD .DC 
= s (s — a) (s — h) (s — c). 
(13) Method of approximating to the square root of 
a non-square number. 
It is a propos of the triangle 7, 8, 9 that Heron gives the 
important statement of his method of approximating to the 
value of a surd, which before the discovery of the passage 
of the Metrica had been a subject of unlimited conjecture 
as bearing on the question how Archimedes obtained his 
approximations to \/3. 
In this case s — 12, s — a = 5, s — h = 4, s — c= 3, so that 
A = 7(12.5.4.3) = P(720). 
Y 2