Take a point E on OB such that BE= 20E, and draw EFH parallel to COD meeting BC, RD in F, H respectively. Let G be the middle point of FH.
Then G is the centre of gravity of the triangle BCD.
Hence, if the angular points B, C be set free and the triangle be suspended by attaching F to E, the triangle will hang in the same position as before, because EFG is a vertical straight line. "For this is proved."
Therefore, as before, there will be equilibrium.
* In Prop. 6 Archimedes takes the separate case in which the angle BCD of the triangle is a right angle so that C coincides with 0 in the figure and F with E. He then proves, in Prop. 7, the same property for the triangle in which BCD is an obtuse angle, by treating the triangle as the difference between two right-angled triangles BOD, BOC and using the result of Prop. 6. I have combined the two propositions in one proof, for the sake of brevity. The same remark applies to the propositions following Props. 6, 7.