A history of Greek mathematics - Wilbourhall.org

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
> BA + AC, we have only to choose G' so that HG', G'L
HG', G'L such 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.
Take any point E on it, and any point D on BE produced
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