78 
ARCHIMEDES 
Hence the centre of gravity of the whole triangle cannot 
but lie on AD. 
It lies, similarly, on either of the other two medians; so 
that it is at the intersection of any two medians (Prop. 14). 
Archimedes gives alternative proofs of a direct character, 
both for the parallelogram and the triangle, depending on the 
postulate that the centres of gravity of similar figures are 
‘ similarly situated ’ in regard to them (Prop. 10 for the 
parallelogram, Props. 11, 12 and part 2 of Prop. 13 for the 
triangle). 
The geometry of Prop. 15 deducing the centre of gravity of 
a trapezium is also interesting. It is proved that, if AD, BG 
are the parallel sides [AD being the smaller), and EF is the 
straight line joining their middle points, the centre of gravity 
is at a point G on EF such that 
GE:GF= (2BG + AD):[2AD + BC). 
Book II of the treatise is entirely devoted to finding the 
centres of gravity of a parabolic segment (Props. 1-8) and 
of a portion of it cut off by a parallel to the base (Props. 9, 10). 
Prop. 1 (really a particular case of I. 6, 7) proves that, if P, P' 
B 
be the areas of two parabolic segments and* D, E their centres 
of gravity, the centre of gravity of both taken together is 
at a point G on DE such that