376 PAPPUS OF ALEXANDRIA 
IV. We now come to the substantive theorem. 
Let FGH be the circle touching all three semicircles (Fig. 4). 
We have then, as in Lemma II, 
BG.BK = BD.BL, 
and for the same reason (regarding FGH as touching the 
semicircles BGC, DUG) 
BC.GL = GD. GK. 
From the first relation we have 
BG: BD = BL : BK, 
whence DC: BD = KL ; BK, and inversely BD: DG=BK: KL, 
while, from the second relation, BG: CD = GK : GL, 
whence BD: DC = KL ; GL. 
Consequently BK: KL = KL : GL, 
or BK. LG = KL 2 . 
But we saw in Lemma II (h) that BK . LG = AM 2 . 
Therefore KL = AM, or = d 1 . 
For the second circle Lemma III gives us 
(Pi + di): d x = p 2 :d 2 , 
whence, since p 1 =z d 1 , p 2 = 2d 2 . 
For the third circle 
[p 2 + d 2 ): d 2 = p> 3 '• d a , 
whence p a = 3d 3 . 
And so on ad infinitum.