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 AB as diameter, C, B any two points on AB, 
and CE, DF being perpendicular to AB, let EF be joined and 
produced, and let BG be drawn perpendicular to EG. To 
prove that 
CB.BD = BG 2 , (1) 
AG.BB = FG 2 , (2) 
AB . BC = EG 2 . (3) 
Join GO, GD, FB, EB, AF. 
(1) Since the angles at G, B are right, F, G, B, B are concyclic. 
Similarly E, G, B, C are concyclic. 
Therefore 
ZBGB=ZBFB 
= Z FAB 
= Z FEB, in the same segment of the semicircle, 
= Z GCB, in the same segment of the circle EGBC. 
And the triangles GCB, BGB also have the angle CBG 
common; therefore they are similar, and GB: BG = BG: BB, 
or CB .BB = BG 2 . 
(2) We have AB . BD = BF 2 ; 
therefore, by subtraction, AC. BB = BF 2 — BG 2 = FG 2 . 
(3) Similarly AB . BC = BE 2 ; 
therefore, by subtraction, from the same result (1), 
AB . BC = BE 2 —BG 2 = EG 2 . 
Thus the lemma gives an extremely elegant construction for 
squares equal to each of the three rectangles.