522 
COMMENTATORS AND BYZANTINES 
(Prop. 33). Prop. 31 preceding these propositions is a par 
ticular case of the constancy of the anharmonic ratio of a 
pencil of four rays. If two sides AB, AG of a triangle meet 
a transversal through D, an external point, in E, F and another 
ray AG between AB and AG cuts DEF in a point G such 
that ED : DF — EG: GF, then any other transversal through 
D meeting AB, AG, AG in K, L, M is also divided harmoni 
cally, i.e. KD : DM = KL: LM. To prove the succeeding pro 
positions, 32 and 33, Serenus uses this proposition and a 
reciprocal of it combined with the harmonic property of the 
pole and polar with reference to an ellipse. 
((3) On the Section of a Gone. 
The treatise On the Section of a Gone is even less important, 
although Serenus claims originality for it. It deals mainly 
with the areas of triangular sections of right or scalene cones 
made by planes passing through the vertex and either through 
the axis or not through the axis, showing when the area of 
a certain triangle of a particular class is a maximum, under 
what conditions two triangles of a class may be equal in area, 
and so on, and solving in some easy cases the problem of 
finding triangular sections of given area. This sort of investi 
gation occupies Props. 1-57 of the work, these propositions 
including various lemmas required for the proofs of the 
substantive theorems. Props. 58-69 constitute a separate 
section of the book dealing with the volumes of right cones 
in relation to their heights, their bases and the areas of the 
triangular sections through the axis. 
The essence of the first portion of the book up to Prop. 57 
is best shown by means of modern notation. We will call h 
the height of a right cone, r the radius of the base; in the 
case of an oblique cone, let be the perpendicular from the 
vertex to the plane of the base, d the distance of the foot of 
this perpendicular from the centre of the base, r the radius 
of the base. 
Consider first the right cone, and let 2x be the base of any 
triangular section through the vertex, while of course 2r is 
the base of the triangular section through the axis. Then, if 
A be the area of the triangular section with base 2x, 
A = x V (r 2 — x 2 + IS).