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 BC. 
Then, if we bisect DC, then bisect the halves, and so on, 
we shall arrive at a length DE less than IH. Divide BC 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 BN, TP, 
FQ lie on AD (Prop. 9); therefore the centre of gravity of the 
A 
figure formed by them all lies on AD; let it be 0. Join OH, 
and produce it to meet in V the parallel through G to AD. 
Now it is easy to see that, if n be the number of parts into 
which DC, AC are divided respectively, 
(sum of small As AMR, MLB... ARN, NUP ...): (A ABC) 
= n.AN 2 :AC 2 
= 1: n ; 
whence 
(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)0H. 
But VH: HO < CE: ED or [n — 1): 1; therefore XH > 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.