A history of Greek mathematics - Wilbourhall.org

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GEMINUS 229
EF, and let the interior angles BEF, EFD be together less
than two right angles.
Take any point H on FD and draw HK parallel to AB
meeting EF in K. Then, if we bisect EF at L, LF at M, MF
at X, 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 i^Q be the same multiple of FO that FE is of i^iY
then shall AB, CD meet in Q.
Let $ be the middle point of FQ and R the middle point of
FS. Draw through R, S, Q respectively the straight lines
RPG, STU, QV parallel to EF. Join MR, LS 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 = FK
there-
And FN, PG in the parallelogram FNPG are equal ;
fore RG = 2FN= FM (whence MR is parallel to FG or AB)
Similarly we prove that SU = 2 FM = FL, and LS is
parallel to FG or AB.
Lastly, by the triangles FLS, QVS, in which the sides FS,
SQ are equal and two angles are respectively equal, Q V = FL.
Therefore QV = LE.
Since then EL, QV are equal and parallel, so are EQ, LV,
and (says Geminus) it follows that AB passes through Q.