A history of Greek mathematics - Wilbourhall.org

DIVISIONS OF FIGURES 339
two opposite sides which the required straight line cuts are
(a) parallel or (b) not parallel. In the first case (a) the
problem reduces to drawing a straight line through E intersecting
the parallel sides in points F, G such that BF+AG
is equal to a given length. In the second case (b) where
BG, AD are not parallel Heron supposes them to meet in H.
The angle at H is then given, and the area ABU. It is then
a question of cutting off from a triangle with vertex H a
triangle HFG of given area by a straight line drawn from E}
which is again a problem in Apollonius's Cutting-ojf of an
area. The auxiliary problem in case (a) is easily solved in
III. 16. Measure AH equal to the given length. Join BH
and bisect it at M. Then EM meets BG, AD in points such
that BF+ AG= the given length. For, by congruent triangles,
BF = GH.
The same problems are solved for the case of any polygon
in III. 14, 15.
A sphere is then divided (III. 17) into segments
such that their surfaces are in a given ratio, by means of
Archimedes, On the Sphere and Cylinder, II. 3, just as, in
III. 23, Prop. 4 of the same Book is used to divide a sphere
into segments having their volumes in a given ratio.
III. 18 is interesting because it recalls an ingenious proposition
in Euclid's book On Divisions. Heron's problem is
'
To divide a given circle into three equal parts by two straight
z 2