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