ARCHIMEDES (Heath) Quadrature of the Parabola
**Proposition 6, 7***
*Suppose a *** ***lever AOB placed horizontally
and supported at its middle point 0. Let a triangle BCD in which the angle
C is right or obtuse be suspended from B and 0, so*** **that C is attached
to 0 and CD is in the same vertical line with 0. Then, if P be such an
area as, when suspended from A, will keep the system in equilibrium,
*P = 1/3 BCD.*
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.

Thus

*P: BCD=OE:AO*
= **1: 3,**
**or**
**P= 1/3 BCD.**
* 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.