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

THE COLLECTION. BOOK IV 373
II. Let (Fig. 2) BC, BD, being in one straight line, be the
any circle as
diameters of two semicircles BGC, BED, and let
FGH touch both semicircles, A being the centre of the circle.
Let M be the foot of the perpendicular from A on BC, r the
radius of the circle FGH. There are two cases according
as BD lies along BC or B lies between D and C\ i.e. in the
first case the two semicircles are the outer and one of the inner
semicircles of the apfi-qXos, while in the second case they are
the two inner semicircles; in the latter case the circle FGH
may either include the two semicircles or be entirely external
to them. Now, says Pappus, it is to be proved that
in case (1) BM:r = (BC+BD) : (BC-BD),
and in case (2) BM : r = (BC-BD) -.'(BC+BD).
374 PAPPUS OF ALEXANDRIA
But
therefore
or
QB.BH=FB.BE\
GB.BK = DB. BL,
BC:BD = BL:BK.
Therefore (BC + BD) : (BC-BD) = (BL + BK) : (BL-BK)
= 2BM:KL.
And
KL = HF=2r;
therefore BM :
It is next proved that BK .
r = (BG + BD) : (BC- BD) .
LG
= AM 2 .
(a)
For, by similar triangles BKH, FLG,
BK:KH=FL.LG, or BK.LG=KH.FL
= AM 2 , (b)
Lastty, since BG :
BD = BL :
BG:GD= BL:KL, or
BK,
from above,
BL.GD = BG.KL
= BC.2r. (c)
Also BD:CD = BK:KL, or BK.GD= BD . KL
= BD.2r. (d)
We will confine ourselves to the first case, represented in
the figure (Fig. 2).
Draw through A the diameter HF parallel to BC. Then,
since the circles BGC, HGF touch at G, and BC, HF are
parallel diameters, GHB, GFC are both straight lines.
Let E be the point of contact of the circles FGH and BED;
then, similarly, BEF, HED are straight lines.
Let HK, FL be drawn perpendicular to BC.
By the similar triangles BGC, BKH we have
BC:BG = BH:BK, or CB .
and by the similar triangles BLF, BED
BF-.BL = BD.BE, or
BK = GB .
BH;
DB.BL = FB.BE.
III. We now (Fig. 3) take any two circles touching the
semicircles BGG, BED and one another.
Let their centres be
A and P, H their point of contact, d, d' their diameters respectively.
Then, if AM, PN are drawn perpendicular to BG,
Pappus proves that
Draw BF perpendicular to
(AM+d):d = PN:d'.
BG and therefore touching the
semicircles BGG, BED at B. Join AP, and produce it to
meet BF in F.
Now, by II. (a) above,
(BG + BD): (BC-BD) = BM:AH,
and for the same reason = BN : PH
it follows that
;
AH:PH=BM: BN
= FA : FP.