THE COLLECTION. BOOK IV 
373 
II. Let (Fig. 2) BC, BD, being in one straight line, be the 
diameters of two semicircles BGC, BED, and let any circle as 
FGII 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 G\ i.e. in the 
first case the two semicircles are the outer and one of the inner 
semicircles of the dp^rjXo?, 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 = (BG—BD) : {BC + BD). 
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, GFG 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 GB . BK = GB. BH; 
and by the similar triangles BLF, BED 
BF: BL = BD: BE, or DB.BL= FB. BE.