GEMINUS 
229 
EF, and let the interior angles BEF, EFD be together less 
than two right angles. 
Take any point II on FI) and draw HK parallel to AB 
meeting EF in K. Then, if we bisect EF at L, LF at M, MF 
at N, and so on, we shall at last have a length, as FN, less 
than FK. Draw FG, NOP parallel to AB. Produce FO to Q, 
and let FQ be the same multiple of FO that FE is of FN; 
then shall AB, CD meet in Q. 
Let B be the middle point of FQ and R the middle point of 
FS. Draw through E, B, Q respectively the straight lines 
RPG, ¡STU, QV parallel to EF. Join MR, LB and produce 
them to T, V. Produce FG to U. 
Then, in the triangles FON, ROP, two angles are equal 
respectively, the vertically opposite angles FON, ROP and 
the alternate angles NFO, PRO; and FO = OR ; therefore 
RP = FN. 
And FN, PG in the parallelogram FNPG are equal; there 
fore RG'= 2FN = FM (wdience MR is parallel to FG or AB). 
Similarly we prove that BU = 2FM = FL, and LB is 
parallel to FG or AB. 
Lastly, by the triangles FLS, QVB, in which the sides FB, 
BQ are equal and two angles are respectively equal, QV — FL. 
Therefore QV — LE. 
Since than EL, QV are equal and parallel, so are EQ, LV, 
and (says Geminus) it follows that AB passes through Q.