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

curve in E 1
, R
THE QUADRATURE OF THE PARABOLA 87
2
, ... . Join
F, QB 2
meeting
l
E l
in F lt
and so on.
QR X
,
and produce it to meet OE in
o Ht h 2 h 3 Hy, a
88 ARCHIMEDES
that is, it takes the trapezium F0 Y
suspended at A to balance
the trapezium E0 1
suspended at H v And P 2
balances E0 X
where it is.
Therefore (FO,) > P 2
.
Similarly (^1^2) > ^2' anc^ so on -
Again AO :0H 1
= E^: X
R X
= (trapezium E 1 2 )
: (trapezium P
2 2),
that is, {R x 2 ) at A will balance (E 1 2 ) suspended at H 1 ,
while P 2
at A balances (Efi^) suspended where it is,
whence P 2
> R Y 2
.
Therefore (F x
0^ > P 2
> (PjOJ,
(P 2 3 ) > P 6
> P 2 3
, and so on;
Now Archimedes has proved in a series of propositions
(6-13) that, if a trapezium such as
1
E 1
E 2 2
is suspended
from H X
H 2
,
and
an area P suspended at J. balances 1E 1
E 2 2
so suspended, it will take a greater area than P suspended at
A to balance the same trapezium suspended from H 2
and
a less area than P to balance the same trapezium suspended
from H . 1
A similar proposition holds with regard to a triangle
such as E n
H n Q suspended where it is and suspended at Q and
H n respectively.
Suppose (Props. 14, 15) the triangle QqE suspended where
it is from OQ, and suppose that the trapezium E0 lt
suspended
where it is, is balanced by an area F^ suspended at A, the
trapezium suspended where El 2 it is, is balanced by P , 2
suspended at A, and so on, and finally the triangle E n O n Q,
suspended where it is, is balanced by P n+1 suspended at A
;
then P Y
+ P 2
+ ... +P n+1 at A balances the whole triangle, so that
P 1
+ P 2
+... + P n+l
= iA_E q Q,
since the whole triangle may be regarded as suspended from
the point on OQ vertically above its centre of gravity.
Now AO:OH
l
= QO:OH l
= Qq:q0 1
= E^O^O^, by Prop. 5,
= (trapezium EO^) : (trapezium
P0 2 ),
and finally,
By addition,
&H n °nQ > P n+i > & R n °nQ'
(R 1 2
)+(R 2
O i ) + ...+(AR n O n
Q)