A history of Greek mathematics - Wilbourhall.org

THE COLLECTION.
BOOK VIII
431
Now, by hypothesis,
whence CA :
GE:EA =BD:DC,
AE = BC : CD,
and, if we halve the antecedents,
therefore AK :
EK
AK:AE= HC:CD;
= HC : HD
or BE :
HD,
whence, componendo, CE :
T^A" = 5i) : DH.
But AF: FB= BD: DC = (J5D : 2)//) . (DH
(1)
:
DC)
= (CE:EK).(DH:DC). (2)
Now, i£XZ) being a transversal cutting the sides of the
triangle KHC, we have
HL:KL = (CE:EK) . (DH
:
DC). (3)
'
[This is Menelaus's theorem ; '
Pappus does not, however,
quote it, but proves the relation ad hoc in an added lemma by
drawing CM parallel to DE to meet HK produced in M. The
proof is easy, for . .
HL LK = ( HL LM) (Zif LK)
.
It follows from (2) and (3) that
= (HD:DC).(CE:EK).]
AF: FB = HL: LK,
and, since AB is parallel to HK, and AH, BK are straight
lines meeting in G, FGL is a straight line.
[This is proved in another easy lemma by reductio ad
absurdum.]