374
PAPPUS OF ALEXANDRIA
But GB.BH = FB.BE:
therefore CB. BK = DB. BL,
or BC:BD = BL : BK\
Therefore {BG + BB) : {BC- BD) = {BL + BK) : {BL - BK)
= 2 BM:KL.
And KL = HF= 2r;
therefore BM:r= {BC + BD): {BG-BD). {a)
It is next proved that BK . LC = AM 2 .
For, by similar triangles BKH, FLC,
BK: KH = FL: LG, or BK. LC = KH, FL
= AM 2 . (6)
Lastly, since BG: BD = BL: BK, from above,
BG: GD = BL: KL, or BL . CD = BG. KL
= BG. 2r. (c)
Also BD: GD = iUf: KL, or Rif .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 respec
tively. Then, if AM, PK are drawn perpendicular to BC,
Pappus proves that
{AM + d):d = PK.d'.
Draw BF perpendicular to 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); {BG- BD) = BM-.AH,
\
and for the same reason = BK: PH;
it follows that AH: PH = BM: BK
= FA : FP.