A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

BG
DB
THE COLLECTION. BOOK VII 409
The algebraical equivalents for Figures (1) and (2) respectively
may be written (if a — AD, b — DC, c = BD,
d = DE)
If ab — cd, then (a±d + c±b) a = (a + c) (a±d),
(a±d + c±b)
b = (c±b) (b + d),
(a±d + c±b) c = (c + a)(c±b),
410 PAPPUS OF ALEXANDRIA
but also what the value is, for three different positions of P in
relation to the four given points.
I will give, as an illustration, the first case, on account of its
elegance. It depends on the following Lemma. AEB being
a semicircle on A B as diameter, C, D any two points on A B,
and GE, DF being perpendicular to A B, let EF be joined and
(a±d + c±b) d— (a±d)(d + b).
Figure (3) gives other varieties of sign. Troubles about
sign can be avoided by measuring all lengths in one direction
from an origin outside the line. Thus, if A = a, OB = b,
&c, the proposition may be as follows
If (d — a) (d — c) = (6 — d) (e-d) and k = e — a + b — c,
then k(d — a) — (b — a)(e — a), k(d — c) = (b — c)(e—c),
Jc(b — d) = (b—a)(b — c)
VIII. Props. 45-56.
More generally, if AD .
then, if
and k(e— d) = (e — a)(e — c).
DC = BD .DE and k = AE±BC,
F be any point on the line, we have, according to the
position of i^in relation to A, B, (7, D, E,
±AF. FC±EF. FB = k. DF.
Algebraically, if 0A = a, OB = b ... OF — x, the equivalent
is: If (d — a) (d — c) — (b — d) (e — d), and k = (e — a) + (b — c),
then
(x — a)(x — c) + (x — e)(b — x) = k (x — d).
By making x — a, b, c, e successively in this equation, we
obtain the results of Props. 41-3 above.
IX. Props. 59-64.
In this group Props. 59, 60, 63 are lemmas required for the
remarkable propositions (61, 62, 64) in which Pappus investigates
singular and minimum ' ' values of the ratio
AP.FDiBP.FC,
where (A, D), (B, C) are point-pairs on a straight line and F
is another point on the straight line. He finds, not only when
the ratio has the singular and minimum ' '
(or maximum) value,
produced, and let BG be drawn perpendicular to EG. To
prove that
Join GG, GD, FB, EB
y
CB.BD = BG 2 , (1)
AG.DB = FG 2 (2)
,
AD . BG = EG 2 . (3)
AF.
(1) Since the angles at G, D are right, F, G, B, D are coneyclic.
Similarly E, G, B, C are concyclic.
Therefore
£BGD = IBFI)
= I FAB
= Z FEB, in the same segment of the semicircle,
= Z GGB, in the same segment of the circle EGBG.
And the triangles GCB, DGB also have the angle GBG
common
;
therefore they are similar, and GB : BG = BG BD,
:
or GB.BD = BG 2 .
(2) We have AB . BD = BF 2 ;
therefore, by subtraction, AG . = BF -BG 2 2 = FG 2 .
= BE 2 ;
(3) Similarly AB .
therefore, by subtraction, from the same result (1),
AD.BG= BE 2 -BG 2 = EG 2 .
Thus the lemma gives an extremely elegant construction for
squares equal to each of the three rectangles.