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.