A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

ZENODORUS 211
Now, by hypothesis, DB + BA = GB + BF;
therefore
DB + BA=HB + BF> HF.
By an easy lemma, since the triangles DEB, ABC are similar,
(DB + BAf = {DL + AKf + (BL + BK) 2
= (DL + AK)* + LK\
Therefore {DL + AKf + LK 2 > HF 2
whence
DL + AK > GL + FK,
and it follows that AF > GD.
But BK > BL; therefore
>{GL + FK) 2 + LK 2 ,
AF.BK > GD.BL.
Hence the hollow-angled (figure) ' ' (KoiXoycoviov) ABFC is
greater than the hollow-angled (figure) GEDB.
Adding A DEB + A BFG to each, we have
h
ADEB + £ABC> AGEB+AFBC.
The above is the only case taken by Zenodorus. The proof
But it fails in the
still holds if EB = BG, so that BK = BL.
case in which EB > BG and the vertex G of the triangle EB
belonging to the non-similar pair is still above D and not
below it (as F is below A in the preceding case). This was
no doubt the reason why Pappus gave a proof intended to
apply to all the cases without distinction. This proof is the
same as the above proof by Zenodorus up to the point where
it is proved that
DL + AK > GL + FK,
but there diverges. Unfortunately the text is bad, and gives
no sufficient indication of the course of the proof ; but it would
seem that Pappus used the relations
and
DL : GL = A DEB : A GEB,
AK : FK = A ABC: A FBC,
AK 2 :DL =A 2 ABC: A DEB,
combined of course with the fact that GB + BF = DB + BA,
in order to prove the proposition that,
according as
DL + AK > or < GL + FK,
ADEB + AABC> or < AGEB + AFBC.
p2
212 SUCCESSORS OF THE GREAT GEOMETERS
The proof of his proposition, whatever it was, Pappus
but in the text as we have it
indicates that he will give later ;
the promise is not fulfilled.
Then follows the proof that the maximum polygon of given
perimeter is both equilateral and
equiangular.
(1) It is equilateral.
For, if not, let two sides of the
maximum polygon, as AB, BC, be
unequal. Join AC, and on iC as
base draw the isosceles triangle AFC
such that AF+ FC = AB + BC. The
area of the triangle AFC is then
greater than the area of the triangle ABC, and the area of
the whole polygon has been increased by the construction:
which is impossible, as by hypothesis the area is a
maximum.
Similarly it can be proved that no other side is unequal
to any other.
(2) It is also equiangular.
For, if
DEC.
possible, let the maximum polygon ABCDE (which
we have proved to be equilateral)
have the angle at B greater than
the angle at D. ThenBA C, DEC are
non-similar isosceles triangles. On
AC, CE as bases describe the two
isosceles triangles FAC, GEC similar
to one another which have the sum
of their perimeters equal to the
sum of the perimeters of BAG,
Then the sum of the areas of the two similar isosceles
triangles is greater than the sum of the areas of the triangles
BAC, DEC) the area of the polygon is therefore increased,
which is contrary to the hypothesis.
Hence no two angles of the polygon can be unequal.
The maximum polygon of given perimeter is therefore both
equilateral and equiangular.
Dealing with the sphere in relation to other solids having