80 
MR. DAVIES’S GEOMETRICAL INVESTIGATIONS 
Let T II be produced to meet S H in S. Then it is obvious from the preceding 
lemmas that the problem is reducible to the description of two circles which shall 
touch the given line II S ; and each divide the lineT U internally and externally in the 
same ratio, or divide it harmonically ; that is, in the one case UC : CT :: UD:DT, 
and in the other TC' : C'U ::TD' : D' U. 
Suppose the points of contact P and P' to be found. Draw M H from M the middle 
of T U perpendicular to T U, and let it meet II S in PI. Draw the lines P O and P' O' 
from the points of contact perpendicular to H S ; then O and O' are the centres of 
the circles. 
First case. Where the line TP is the ante- 
cedent, and the ratio the least possible. 
By Leslie’s G eom., vi. 7- MT 2 = MC.MD, 
that is, 
MT 2 = (MS-SC) (MS-SD) 
= M S 2 -MS(SC + SD)+SD.SC 
= M S 2 - 2 MS . SO + SD . S C. 
But by the similar triangles SPO, SHM, 
MS.SO = PS.SH; 
and by the circle S C . S D = S P 2 . 
Hence 
M T 2 = M S 2 - 2 H S . S P + S P 2 
= HS 2 - 2HS.SP + SP 2 -HM 2 . 
Hence 
(HS-SP) 2 = MT 2 + MH 2 = HT 2 . 
That is S P = H S — H T. 
Hence this 
Construction . — Draw the perpendicular M H (from M) to T U, meeting H S in H. 
With centre II and distance H T or PI U describe a circle cutting HS in P and P'. 
These are the points at which the ratios are those sought, as is too evident from the 
analysis to need a formal demonstration. 
The ratio T P : P U is hence the least, and T P' : P' U the greatest that lines drawn 
from T and U to meet in the line H S, can possibly have ; or the greatest and least, 
if the order of the terms of the ratio be changed. At H, midway between P and P', 
they have a ratio of equality. 
Lemma IV. T heorem. — A of, more than two pairs of lines can be inflected from the 
same two points T and U to meet in the same straight line PI S, and have to one another 
a given ratio, the order of the terms of the ratio being given. 
For since the locus of all the intersections of all lines which can be so drawn is a 
circle, (Lemma 1.) and a circle can cut a straight line in only two points, the truth 
of the proposition follows. 
Secoiid case. Where the line T P' is the an- 
tecedent, and the ratio the least possible. 
By Leslie’s Geom., ib. MU 2 = MC'.MD', 
that is, 
MU 2 =(SC'-SM) (SD'-S M) 
= SC'.SD'-SM(SC'+SD')-fSM 2 
==SC'.SD'-2SM.SO'+SM 2 . 
But by the similar triangles SP'O', SHM, 
MS.SO' = P' S.SH; 
and by the circle S C' . S D' = S P' 2 . 
Hence 
MU 2 = SP' 2 - 2HS.SP' + SM 2 
= SP' 2 — 2HS.SP' + HS 2 — HM 2 . 
Hence 
(SP'-HS) 2 =MU 2 + MH 2 =HU 2 =HT 2 . 
Thatis SP' = HS + HT. 
