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that is, it takes the trapezium FO l suspended at A to balance 
the trapezium E0 X suspended at H v And 1\ balances E0 X 
where it is. 
Therefore (P0 X ) > 1\. 
Similarly (^1^2) > P 2 > anc ^ 80 on - 
Again AO : 0H X = E x 0 X : 0 1 R l 
= (trapezium E x 0.f): (trapezium R x 0 2 ), 
that is, (Ri0 2 ) at A will balance (E x 0 2 ) suspended at H x , 
while P 2 at A balances (E x 0 2 ) suspended where it is, 
whence P 2 > R x 0 2 . 
Therefore (^1^2) > P 2 > i^i0 2 ), 
(P 2 0 3 ) > P 3 > R 2 0 3 , and so on; 
and finally, AE n O n Q > P n+X > AR n O n Q. 
By addition, 
{RM + {R 2 0 3 ) + ... + (AR n O n Q) < P 2 + P 3 + ... +P n+} ; 
therefore, a fortiori, 
(-^1^2) + (^2^3) + ••• + AP n O n Q < P x + P 2 +... +P n+1 
<{F0 x ) + {F x 0 2 )+... + AE n 0 n Q. • 
That is to say, we have an inscribed figure consisting of 
trapezia and a triangle which is less, and a circumscribed 
figure composed in the same way which is greater, than 
P x +P 2 +... + P n+1 , i.e. §AEqQ. 
It is therefore inferred, and proved by the method of ex 
haustion, that the segment itself is equal to j A EqQ (Prop. 16). 
In order to enable the method to be applied, it has only 
to be proved that, by increasing the number of parts in Qq 
sufficiently, the difference between the circumscribed and 
inscribed figures can be made as small as we' please. This 
can be seen thus. We have first to show that all the parts, as 
qF, into which qE is divided are equal. 
We have E 1 0 1 : 0 X R X = QO : 0H X — {n+ l): 1, 
or 0 X R X = ——.E X 0 X , whence also 0 2 S = ——.0 2 E 2 . 
1 1 n +\ 1 i’ 2 n+l 2 2 
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