ARCHIMEDES 
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proof's. We do not find him here assuming, as in The Method, 
that, if all the" lines that can be drawn in a figure parallel to 
(and including) one side have their middle points in a straight 
line, the centre of gravity must lie somewhere on that straight 
line; he is not content to regard the figure as made up of an 
infinity of such parallel lines; pure geometry realizes that 
the parallelogram is made up of elementary parallelograms, 
indefinitely narrow if you please, but still parallelograms, and 
the triangle of elementary trapezia, not straight lines, so 
that to assume directly that the centre of gravity lies on the 
straight line bisecting the parallelograms would really be 
a petitio principii. Accordingly the result, no doubt dis 
covered in the informal way, is clinched by a proof by reductio 
ad absurdum in each case. In the case of the parallelogram 
ABGD (Prop. 9), if the centre of gravity is not on the straight 
line EF bisecting two opposite sides, let it be at H. Draw 
HK parallel to AD. Then it is possible by bisecting AE, ED, 
then bisecting the halves, and so on, ultimately to reach 
a length less than KH. Let this be done, and through the 
points of division of AD draw parallels to MR or DC making 
a number of equal and similar parallelograms as in the figure. 
The centre of gravity of each of these parallelograms is 
similarly situated with regard to it. Hence we have a number 
of equal magnitudes with their centres of gravity at equal 
distances along a straight line. Therefore the centre of 
gravity of the whole is on the line joining the centres of gravity 
of the two middle parallelograms (Prop. 5, Cor. 2). But this 
is impossible, because H is outside those parallelograms. 
Therefore the centre of gravity cannot but lie on EF. 
Similarly the centre of gravity lies on the straight line 
bisecting the other opposite sides AB, CD; therefore it lies at 
the intersection of this line with EF, i.e. at the point of 
intersection of the diagonals.