A history of Greek mathematics - Wilbourhall.org

432 PAPPUS OF ALEXANDRIA
We have next to prove that
EL = LB.
'
Now [again by
drawing GN parallel to HK to meet ED produced in N~\
But, by (1) above,
therefore GK :
Menelaus's theorem ',
proved ad hoc by
EL:LD = (EK : KG)
.
(CH
:
CE.EK = BD:DH;
HD = CH :
KE = BH :
HD). (4)
HD,
so that (EK:KC).(CH:HD)=1, and therefore, from (4),
EL = LD.
It remains to prove that FG = 2GL, which is obvious by
parallels, since FG :
GL
= AG : GH
Two more propositions follow
=2:1.
with reference to the centre
of gravity. The first is, Given a rectangle with AB, BG as
adjacent sides, to draw from C a straight line meeting the side
opposite BC in a point D such that, if the trapezium ADCB is
hung from the point D, it will rest with AD, BG horizontal.
G
E
B l\A I N C
In other words, the centre of gravity must be in DL, drawn
perpendicular to BG. Pappus proves by analysis that
GIj 2 = 3BL 2 , so that the problem is reduced to that of
dividing BG into parts BL, LG such that this relation holds.
The latter problem is solved (Prop. 6) by taking a point,
say X, in GB such that GX = 3 XB, describing a semicircle on
BG as diameter and drawing XY at right angles to BG to
meet the semicircle in F, so that XY' i = ^s BG 2 ,
and then
dividing GB at X so that
GL :LB = CX: XY(= i
:
J*/ 3 = x/3 :
1).
The second proposition is this (Prop. 7).
Given two straight
lines AB, AG, and B a fixed point on AB, if GD be drawn