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,