A history of Greek mathematics - Wilbourhall.org

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PROOF OF THE FORMULA OF HERON' 323
Draw OL at right angles to OC cutting BC in K, and BL at
right angles to BC meeting OL in L.
Then, since
a quadrilateral in a circle.
Join GL.
each of the angles COL, CBL is right, COBL is
Therefore Z COB + Z 0X5 =25.
But LCOB + LAOF= 25, because 40, BO, CO bisect the
angles round 0, and the angles GOB, AOF are together equal
to the angles 400, 50.F, while the sum of all four angles
is equal to 45.
Consequently
AA0F = Z CLB.
Therefore the right-angled triangles AOF, CLB are similar
therefore
BC:BL = AF:F0
= BH-.OD,
and, alternately, CB:BH = BL: OD
= BK:KD;
whence, componendo, GH:HB — BD : DK.
It follows that
CH :CH.HB = BD.DC 2 : CD. DK
= BD.DC: OD 2 ,
since the angle COK is right.
Therefore (A ABC) 2 = CH 2 . OD
2 (from above)
= CH.HB. BD.DC
= s(s — a) (s - h) (s — e).
(/3) 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 VS.
In this case s = 12, s — a = 5, s — fr = 4, s — c = 3, so that
A = /(12 .5.4.3) = 7(720).
y2