A history of Greek mathematics - Wilbourhall.org

SERENUS 525
the base of the cone to the projection of the vertex on its
plane ; the areas of the axial triangles are therefore proportional
to the generators of the cone with the said circle as
base and the same vertex as the original cone.
Prop. 50 is to
the effect that, if the axis of the cone is equal to the radius of
the base, the least axial triangle is a mean proportional
between the greatest axial triangle and the isosceles triangular
section perpendicular to the base ;
that is, with the above notation,
if r = V(p 2 + d 2 ), then r \/{p 2 + d 2 ) :rp = rp:p b^.od,
then a + d < b + c (a,d are the sides other than the base of one
axial triangle, and b, c those of the other axial triangle compared
with it; and if ABC, ADEbe two axial triangles and
the centre of the base, BA 2 + AC 2 =DA 2 + AE 2
because each
of these sums is equal to 2 A 2 + 2 BO 2 ,
Prop. 1 7). This proposition
again depends on the lemma (Props. 52, 53) that, if
straight lines be ' inflected ' from the ends of the base of
a segment of a circle to the curve (i. e. if we join the ends
of the base to any point on the curve) the line (i. e. the sum of
the chords) is greatest when the point taken is the middle
point of the arc, and diminishes as the point is taken farther
and farther from that point.
Let B be the middle point of the
arc of the segment ABC, D, E any
other points on the curve towards
G\ I say that
AB + BC>AD + DG>AE+EC.
With B as centre and BA as radius
describe a circle, and produce AB,
AD, AE to meet this circle in F, G,
H. Join FG, GC, HG
Since AB = BG = BF, we have AF = AB + BG
Also the
angles BFC, BGF are equal, and each of them is half of
the angle ABG.