—_— CO  _——  ———— 
Strong’s Problems. 63 
A draw any line AC, cutting the circles in C, H, N, O. The 
segments CBH, NPO are similar, and likewise the remain- 
ing segments CDH, 
Demonst. For draw FB, GP to the points of contact of 
"OF. RG, : and LCE AK: QB : PR. pera 
fore Or. RG: oR; alternately QF: BQ:: 
PR; by Comp. FB: QF :: PG: RG, that is, FH: Ors: 
GO: GR ; (substituting for F B and PG their equals FH 
and GO.) Now ve an ae Pat _ are each of them 
less than a right a cs less than a semi- 
circle) hemare wel VL 7 7) ie a angles FQH and pt 
being eq angles FQ e similar, and th 
angles FH, RGO are equal. In tke manner it may be 
shown that the angles CFQ, RGN are equal. Whence the 
angle CFH = angle NGO. Therefore their halves CDH, 
NEO will likewise be equal. Therefore the segments CDH, 
NEO are similar, and likewise the segments “CBH, NPO. 
(Euc. Def. B. 3.) Wherefore A is the point required. 
Case a When the point falls within the two circles. 
wg. 16. Const. Let AF'N, HBK, be the two given cir- 
cles, a (Prob. 7. C. 2.) the tangent BA cutting the 
line DE (which join the centres of the given circles) in C. 
Then will C be the point required. 
Demonst. For through C, draw any line FCH, cutting 
the circles in F, G, HL, L Join EA, DB, which being per- 
pendicular to AB, are ‘parallel to each other. The angles 
LCE, DCM being vertical are equal. For the same reason 
ACT angle MCB. Therefore the triangles ACL, MCB, 
as also the triangles LCE, DCM are similar. Therefore 
AL:MB::LC:CM wid LE:MD::LC: MC, whence 
by equality, me MB::LE:MD; slcoceeti ly. AL:MB: 
LE: compos. EA or EG: LE: :DB or DI: DM. 
Now the ais LGE, DIM are each of them less than a 
right angle ; Sitshore (Eue. VI. 7. ) the triangles LEG, 
IM are similar, and the angle LEG = angle IDM. In 
like manner it may be shown that the angle FEL = == angle 
MDH. Therefore the whole angle IDH == whole angle 
