418
PAPPUS OF ALEXANDRIA
This is equivalent to the general relation between four
points on a straight line discovered by Simson and therefore
wrongly known as Stewart’s theorem:
AD 2 . BC+ BD 2 . CA + CD 2 . AB + BG. CA . AB = 0.
(Simson discovered this theorem for the more general case
where D is a point outside the line ABC.)
An algebraical equivalent is the identity
{d—a) 2 (b — c) + {d — b) 2 {c — a) + {d — c) 2 (a — h)
+ {b—c) (c — a) (a — b) = 0.
Pappus’s proof of the last-mentioned lemma is perhaps
worth giving.
F
1 , ,
A CD 8
C, D being two points on the straight line AB, take the
point F on it such that
FD : DB = AC: CB. (1)
Then FB : BD = AB : BG,
and {AB — FB) : {BG—BD) — AB : BG,
or AF: CD = AB : BG,
and therefore
AF. CD : CD 2 = AB: BG. (2)
From (1) we derive
A G
~ • DB 2 = FD . DB,
L/Jj
and from (2)
A R
~ • CD 2 = AF. CD.
JjL
We have now to prove that
AD 2 + BD . DF = AC 2 + AG. CB + AF. CD,
or AD 2 + BD . DF = CA .AB+AF.CD,