A history of Greek mathematics - Wilbourhall.org

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)
And
KL = HF=2r;
therefore BM: r = (BG + BD) :
It is next proved that BK .
LG
= AM 2 .
= 2BM:KL.
(BC- BD) .
(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)
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.