Archimedes’ “On the Equilibrium of Planes”, Prop 10

 

by Corry Shores
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Archimedes

On the Equilibrium of Planes

or

The Centres of Gravity of Planes,

Book I

Proposition 10

[The following is quotation]

 

Proposition 10

P10. The centre of gravity of a parallelogram is the point of intersection of its diagonals.

For, by the last proposition, the centre of gravity lies on each of the lines which bisect opposite sides. Therefore it is at the point of their intersection; and this is also the point of intersection of the diagonals.

Alternative proof

Let ABCD be the given parallelogram, and BD a diagonal. Then the triangles ABD, CDB are equal and similar, so that [Post. 4], if one be applied to the other, their centres of gravity will fall one upon the other.

(Heath 195)

Suppose F to be the centre of gravity of the triangle ABD. Let G be the middle point of BD. Join FG and produce it to H, so that FG = GH.

If we then apply the triangle ABD to the triangle CDB so that AD falls on CB and AB on CD, the point F will fall on H.

But [by Post. 4] F will fall on the centre of gravity of CDB. Therefore H is the centre of gravity of CDB.

Hence, since F, H are the centres of gravity of the two equal triangles, the centre of gravity of the whole parallelogram is at the middle point of FH, i.e. at the middle point of BD, which is the intersection of the two diagonals.

(Heath 196)

 

From:

Archimedes. “On the Equilibrium of Planes or The Centres of Gravity of Planes, Book I”. In The Works of Archimedes. Ed. T.L. Heath. Cambridge UP, 1897. Obtained at

https://archive.org/details/worksofarchimede00arch