A history of Greek mathematics - Wilbourhall.org

ON PLANE EQUILIBRIUMS, I 77
The proof in the case of the triangle is similar (Prop. 13).
Let AD be the median through A. The centre of gravity
must lie on AD.
For, if not, let it be at H, and draw HI parallel to BG.
Then, if we bisect DC, then bisect the halves, and so on,
we shall arrive at a length DE less than IH. Divide BG into
lengths equal to DE, draw parallels to DA through the points
of division, and complete the small parallelograms as shown in
the figure.
The centres of gravity of the whole parallelograms SN, TP,
FQ lie on AD (Prop. 9)
; therefore the centre of gravity of the
Join OH,
figure formed by them all lies on AD; let it be 0.
and produce it to meet in Fthe parallel through G to AD.
Now it is easy to see that, if n be the number of parts into
which DG, AC are divided respectively,
(sum of small As AMR, MLS... ARN, NUP ...)
: (A ABC)
whence
= n. AN 2 :AC 2
= 1 : n
;
(sum of small As) : (sum of parallelograms) = 1 : (n—
1).
Therefore the centre of gravity of the figure made up of all
the small triangles is at a point X on OH produced such that
XH=(n-l)OH.
But VH: HO VH.
It follows that the centre of gravity of all the small
triangles taken together lies at X notwithstanding that all
the triangles lie on one side of the parallel to AD drawn
through X : which is impossible.