A history of Greek mathematics - Wilbourhall.org

ZENODORUS 211
Now, by hypothesis, DB + BA = GB + BF;
therefore
DB + BA=HB + BF> HF.
By an easy lemma, since the triangles DEB, ABC are similar,
(DB + BAf = {DL + AKf + (BL + BK) 2
= (DL + AK)* + LK\
Therefore {DL + AKf + LK 2 > HF 2
whence
and it follows that AF > GD.
But BK > BL; therefore
>{GL + FK) 2 + LK 2 ,
DL + AK > GL + FK,
AF.BK > GD.BL.
Hence the hollow-angled (figure) ' ' (KoiXoycoviov) ABFC is
greater than the hollow-angled (figure) GEDB.
Adding A DEB + A BFG to each, we have
h
ADEB + £ABC> AGEB+AFBC.
The above is the only case taken by Zenodorus. The proof
But it fails in the
still holds if EB = BG, so that BK = BL.
case in which EB > BG and the vertex G of the triangle EB
belonging to the non-similar pair is still above D and not
below it (as F is below A in the preceding case). This was
no doubt the reason why Pappus gave a proof intended to
apply to all the cases without distinction. This proof is the
same as the above proof by Zenodorus up to the point where
it is proved that
DL + AK > GL + FK,
but there diverges. Unfortunately the text is bad, and gives
no sufficient indication of the course of the proof ; but it would
seem that Pappus used the relations
and
DL : GL = A DEB : A GEB,
AK : FK = A ABC: A FBC,
AK 2 :DL =A 2 ABC: A DEB,
combined of course with the fact that GB + BF = DB + BA,
in order to prove the proposition that,
according as
DL + AK > or < GL + FK,
ADEB + AABC> or < AGEB + AFBC.
p2