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

ZENODORUS 209
The triangles NMK, HDL are now equal in all respects, and
NK is equal to HL, so that GK < HL.
But the area of the polygon ABC is half the rectangle
contained by GK and the perimeter, while the area of the
polygon DEF is half the rectangle contained by HL and
the same perimeter.
is the greater.
Therefore the area of the polygon DEF
(2) The proof that a circle is greater than any regular
polygon with the same perimeter is deduced immediately from
Archimedes's proposition that the area of a circle is equal
to the right-angled triangle with perpendicular side
equal to
the radius and base equal to the perimeter of the circle
Zenodorus inserts a proof in extenso of Archimedes's proposition,
with preliminary lemma. The perpendicular from
the centre of the circle circumscribing the polygon is easily
proved to be less than the radius of the given circle with
perimeter equal to that of the polygon ; whence the proposition
follows.
(3) The proof of this proposition depends on some preliminary
lemmas. The first proves that, if there be two
triangles on the same base and with the
same perimeter, one being isosceles and
the other scalene, the isosceles
has the greater area.
triangle
(Given the scalene
triangle BDC on the base BC, it is easy to
draw on BC as base the isosceles triangle
having the same perimeter. We have
only to take BH equal to ±(BD + DC),
bisect BC at E, and erect at E the perpendicular
AE such that AE = BH 2
2 -BE\)
Produce BA to F so that BA — AF, and join AD, DF.
Then BD + DF> BF, i.e. BA + AC, i.e. BD + DC, by hypothesis;
therefore DF > DC, whence in the triangles FAD,
CAD the angle FAD > the angle CAD.
Therefore
Z FA D > \ Z FA C
> LBCA.
Make the angle FAG equal to the angle BCA or ABC, so
that AG is parallel to BC; let BD produced meet AG in G,
and join GG.
1523.2 P
210 SUCCESSORS OF THE GREAT GEOMETERS
Then
The second lemma is
A ABC = A GBC
> ADBC.
to the effect that, given two isosceles
triangles not similar to one another, if we construct on the
same bases two triangles similar to one another such that the
sum of their perimeters is equal to the sum of the perimeters
of the first two triangles, then the sum of the areas of the
similar triangles is greater than the sum of the areas of
the non-similar triangles. (The easy construction of the
similar triangles is given in a separate lemma.)
Let the bases of the isosceles triangles, EB, BC\ be placed in
one straight line, BG being greater than EB.
Let ABC, DEB be the similar isosceles triangles, and FBG,
GEB the non-similar, the triangles being such that
BA + AC + ED +DB = BF+ FG+EG + GB.
Produce AF, GD to meet the bases in K, L.
AK, GL bisect BG, EB at right angles at K, L.
Produce GL to H, making LH equal to GL.
Join HB and produce it to J\ r join HF.
;
Then clearly
Now, since the triangles A BG, DEB are similar, the angle
ABC is equal to the angle DEB or DBE.
Therefore Z NBG ( = Z HBE = Z GBE) > Z DBE or Z ABC
;
therefore the angle ABH
f
and a fortiori the angle FBH, is
less than two right angles, and HF meets BK in some point M.