THE COLLECTION. BOOK III
367
Now EA + AC > EF+ FC
> EG + GG and > GG, a fortiori.
Produce GC to K so that GK = EA + AG, and with G as
centre and GK as radius describe a circle. This circle will
meet HG and HG, because GH — EB > BI) or DA +AC and
> GK, a fortiori.
Then HG + GL = BE+EA +AG = BA + A G:
To obtain two straight lines HG', G'Lj such that HG' + G'L
>BA + AG, we have only to choose G' so that HG', G'L
enclose the straight lines HG, GL completely.
Next suppose that, given a triangle ABC in whidh BG > BA
A
> AG, we are required to draw from two points on BG 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 BG and BD because
BA > AG, 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 I/ on BH can be found such that DLL is equal
to A B by marking off DN on DB equal to A B and drawing
with D as centre and DN as radius a circle meeting BH
in 7/. Also, if DH be joined, DH —AG.
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