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

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
VS.
In this case s = 12, s — a = 5, s — fr = 4, s — c = 3, so that
approximations to
A = /(12 .5.4.3) = 7(720).
y2
324 HERON OF ALEXANDRIA
'Since', says Heron, 1 '
720 has not its side rational, we can
obtain its side within a very small difference as follows. Since
the next succeeding square number is 729, which has 27 for
its side, divide 720 by 27. This gives 26|. Add 27 to this,
making 53§, and take half of this or 26 J J.
The side of 720
will therefore be very nearly 26| §. In fact, if we multiply
26J§ by itself , the product is 720^, so that the difference (in
the square) is .
'
^
If we desire to make the difference still smaller than 3^-, we
shall take 720^ instead of 729 [or rather we should take
26J-| instead of 27], and by proceeding in the same way we
shall find that the resulting difference is much less than £$'
In other words, if we have a non-square number A, and a 2
is the nearest square number to it, so that A = a 2 + b,
have, as the first approximation to */A.
for a second approximation we take
and so on. 2
1
Metrica, i. 8, pp. 18. 22-20. 5.
then we
«!=!(«+ -); (D
2
The method indicated by Heron was known to Barlaam and Nicolas
Rhabdas in th'e fourteenth century. The equivalent of it was used by
Luca Paciuolo (fifteenth -sixteenth century), and it was known to the other
Italian algebraists of the sixteenth century. Thus Luca Paciuolo gave
H
2\i 2^ and 2 T 9^1n as successive approximations to */6. He obtained
2
«. a 01 my-s
2
the first as 2+ n—x, *
^—7;, the second as 2| —
l
rtl
2.2'
J ^, and
, the third as
2 t\
tt-irir- The above rule - ives l(2+i) = 2|, i(|+-^)-2A,
" • •
20
1 fiilj. l_liP\ 9JL«JL
2 \20^ 48/ — ^1!>60-
The formula of Heron was again put forward, in modern times, by
Buzengeiger as a means of accounting for the Archimedean approximation
to \/3, apparently without knowing its previous history. Bertrand
also stated it in a treatise on arithmetic (1853-). The method, too, by
which Oppermann and Alexeieff sought to account for Archimedes's
approximations is in reality the same. The latter method depends on
the formula
i(a + /3): v / (a3)-yW): a
2^-
Alexeieff separated A into two factors a , b ,
and pointed out that if. say.
"o> \/^4 >?? ,
then, i(a +b )>^/A> ** or -\ °