DIVISIONS OF FIGURES 
339 
two opposite sides which the required straight line cuts are 
(a) parallel or (?;) not parallel. In the first case [a) the 
problem reduces to drawing a straight line through E inter 
secting the parallel sides in points F, G such that BF+AG 
E 
is equal to a given length. In the second case (6) where 
BG, AD are not parallel Heron supposes them to meet in H. 
The angle at H is then given, and the area ABB. 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 Gutting-off 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 pro 
position in Euclid’s book On Divisions. Heron’s problem is 
‘ To divide a given circle into three equal parts by two straight 
z 2