14 
ARISTARCHUS OF SAMOS 
Prop. 14 (Fig. 3), 
The arcs OM, ML, LP, PN are all equal; therefore so are 
the chords. BM, BP are tangents to the circle MQP, so that 
CM is perpendicular to BM, while BM is perpendicular to OL. 
(Prop. 13) 
(see Prop. 11) 
(Prop. 13) 
therefore, by similar triangles, if YO, ZN meet in X, 
AX : XR >9:1, 
and convertendo, XA :AR< 9:8. 
But AB > 18 BC, and, a fortiori, > 18 BR, 
whence AB > 18 (AR — AB), or 19 AB'> 18 AR; 
that is, AR:AB< 19:18. 
Therefore, ex aequali, 
XA:AB < 19 : 16, 
and, convertendo, AX :XB > 19:3; 
therefore (diam. of sun) : (diam. of earth) > 19:3. 
Lastly, since GB:GR> 675:1, • (Prop. 14) 
GB : BR < 675:674. 
Therefore the triangles LOS, GMR are similar. 
Therefore SO : MR = SL: RC. 
But SO < 2 MR, since ON < 2 MP ; 
therefore SL < 2 RG, 
and, a fortiori, SR < 2 RC, or SC < 3 RC, 
that is, CR : CS >1:3. 
Again, MG :CR = BG: CM 
>45:1; 
therefore, ex aequali, 
CM: GS >15:1. 
And BG: CM >45:1; 
therefore BG: CS > 675:1. 
Prop. 15 (Fig. 1). 
We have NO : (diam. of sun) <1:9, 
and, a fortiori, YZ: NO >9:1;