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, NEO. 
Demonst. For draw FB, GP to the points of contact of 
the tangent, and they will be perpendicular to it and conse- 
quently parallel to each other; draw also FC, FH, GN, 
GO. And suppose the line AC cuts BF, PG in Q, R. 
Now BF, PG being parallel, the triangles ABF, APG, as 
also the triangles AQF, ARG are similar. Whence— 
AQ:AK:: QF: RG, and AQ: AK: : QB: PR. There- 
fore QF: RG:: QB: PR; alternately QF :BQ:: RG: 
PR; by Comp. FB: QF: : PG: RG, that is, FH: QF:: 
GO: GR; (substituting for FB and PG ther equals FH 
and GO.) Now the angles FHQ, GOR are each of them 
less than aright angle (standing on arcs less than a semi- 
circle) wherefore (Eucl. VI. 7) the angles FQH and GRO 
being equal, the triangles FQH, GRO are similar, and the 
angles QFH, RGO are equal. In like manner it may be 
shown that the angles CFQ, RGN are equal. Whence the 
angle CKH = 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 II. When the point falls within the two circles. 
Fig. 16. Const. Let AFN, HBK, be the two given cir- 
cles. Draw (Prob. 7. C. 2.) the tangent BA cutting the 
line DE (which join the centres of the given circles) mm ©. 
Then will C be the point required. 
Demonst. For through C, draw any line FCH, cutting 
the circlesin F, G, H, I. Joi 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 
ACL. = angle MCB. Therefore the triangles ACL, MCB, 
as also the triangles LCE, DCM are similar. Therefore 
Al,:MB::LC >CM and LE: MD::LC: MC, whence 
by equality, AL: MB::LE:MD; alternately, AL: MB:: 
LE:MD; by compos. EA or EG: LE: : DB or DI: DM. 
Now the angles LGE, DIM are each of them less than a 
tight angle; therefore (Euc. VI. 7.) the triangles LEG, 
DIM 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 
