A history of Greek mathematics - Wilbourhall.org

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
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 . DB = BF -BG 2 2 = FG 2 .
BG = 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.