A history of Greek mathematics - Wilbourhall.org

76 ARCHIMEDES
proofs.
We do not find him here assuming, as in The Method,
that, if all the lines that can be drawn in a figure parallel
(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 i^etitio principii. Accordingly the result, no doubt discovered
in the informal way, is clinched by a proof by reductio
ad absardum in each case. In the case of the parallelogram
ABCD (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
to
points of division of AD draw parallels to AB 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.