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

Now
THE COLLECTION. BOOK III 367
EA+AC > EF+FC
> EG + GC and > GC, a fortiori.
Produce GC to K so that GK = EA+AC, and with G as
centre and GK as radius describe a circle. This circle w T ill
meet EC and HG, because GH = EB > BD or DA+AC and
> GK, a fortiori.
Then HG + GL = BE+EA+AC=BA + AC.
To obtain two straight lines HG', G'L such that HG'+G'L
> BA + AC, we have only to choose G' so that HG', G'L
enclose the straight lines HG, GL completely.
Next suppose that, given a triangle A BC in which BC > BA
> AC, we are required to draw from two points on BC to
an internal point two straight lines greater respectively than
BA, AC.
With B as centre and BA as radius describe the arc AEF.
and any point D on BE produced
Take any point E on it,
but within the triangle. Join DC, and produce it to G so
that DG = AC. Then with D as centre and DG as radius
describe a circle. This will meet both BC and BD because
BA > AC, and a fortiori DB > DG.
Then, if L be any point on BH, it is clear that BD, DL
are two straight lines satisfying the conditions.
A point L' on BH can be found such that DL' is equal
to A B by marking off DN on DB equal to A B and drawing
with D as centre and DiV as radius a circle meeting BH
in L'. Also, if DH be joined, DH = AC.
Propositions follow (35-9) having a similar relation to the
Postulate in Archimedes, On the Sphere and Cylinder, I,
about conterminous broken lines one of which wholly encloses
368 PAPPUS OF ALEXANDRIA
the other, i.e. it is shown that broken lines, consisting of
several straight lines, can be drawn with two points on the
base of a triangle or parallelogram as extremities, and of
greater total length than the remaining two sides of the
triangle or three sides of the parallelogram.
Props. 40-2 show that triangles or parallelograms can be
constructed with sides respectively greater than those of a given
triangle or parallelogram but having a less area.
Section (4). The inscribing of the five regular solids
in a
sphere.
The fourth section of Book III (pp. 132-62) solves the
problems of inscribing each of the five regular solids in a
given sphere. After some preliminary lemmas (Props. 43-53),
Pappus attacks the substantive problems (Props. 54-8), using
the method of analysis followed by synthesis in the case of
each solid.
(a) In order to inscribe a regular pyramid or tetrahedron in
the sphere, he finds two circular sections equal and parallel
to one another, each of which contains one of two opposite
edges as its diameter.
If d be the diameter of the sphere, the
parallel circular sections have d' as diameter, where d 2 — \d' 2 .
(b) In the case of the cube Pappus again finds two parallel
circular sections with diameter df such that d = 2 ^d' 2 ;
a square
inscribed in one of these circles is one face of the cube and
the square with sides parallel to those of the first square
inscribed in the second circle is the opposite face.
(c) In the case of the octahedron the same two parallel circular
sections with diameter d' such that d = 2 fcT
2
are used; an
equilateral triangle inscribed in one circle is
one face, and the
opposite face is an equilateral triangle inscribed in the other
circle but placed in exactly the opposite way.
(d)
In the case of the icosahedron Pappus finds four parallel
circular sections each passing through three of the vertices of
the icosahedron ; two of these are small circles circumscribing
two opposite triangular faces respectively, and the other two
circles are between these two circles, parallel to them, and
equal to one another. The pairs of circles are determined in