564 
PROBLEMS AND SOLUTIONS. 
[383 
Consider now the points A, K as fixed points on the conic; then, revolving about 
A the constant angle DAB, and about K the constant (zero) angle DKB, the locus 
of B is (by the theorem of the anharmonic relation of the points of a conic) given 
in the first instance as a conic through the points A, K; but, observing that 
a position of the angle DAB is TAK, and that the corresponding position of DKB 
is AKA, the line AK is part of the locus; and the locus is made up of this line 
and a line BG. And, conversely, given the fixed points A, K, and the line BC, the 
original conic is, by Newton’s theorem, described by means of the constant angles 
DAB, DKB revolving about these points in such a manner that the arms AB, KB 
generate by their intersections the line BG. This being so, the other two arms 
AD, KD generate by their intersections the conic. 
And then, considering the two positions DAB, EAG of the angle DAB (so that 
D, B are in a line with K, and E, G are also in a line with K), we have 
Z DAB = Z EAG, that is, Z DAE = Z BAG, which is Mr Clifford’s theorem. 
It has been seen that, A being given, the same line BG is obtained whatever 
be the position of the point K\ and, taking AK for the normal at A, it at once 
appears geometrically that (as remarked by Mr Clifford) the line BG is the polar of 
the point © of intersection of all the chords which subtend a right angle at A. 
{Professor Cayley’s lemma may be otherwise proved, as follows: 
The trilinear equation of the conic, referred to two tangents (a at A, /3 at S) 
and their chord of contact (y or ZL$), is U= \a(3 — y 2 = 0; and the equation of two 
straight lines (AK, AD) equally inclined to a, y is 
(a — ^y) (yaa — y) = 0, or V = a 2 + y 2 — (fx + /jt 1 ) ay = 0; 
also U +V= 0 denotes a conic passing through the intersections of U and V; but 
U + V is resolvable into a = 0, or the tangent AT, and 
a + X/3 — (/x + ¡jt 1 ) y = 0, 
which is, therefore, the equation of the chord KD : whence we see that KD meets 
AS (or y) in a point B (given by y = 0, a + \/3 = 0) whose position is independent of 
¡x, that is, of the equal angles SAD, TAK.} 
[Voi. i. pp. 125—127.] 
1478. (By J. McDowell, M.A.)—(a) Two sides of a given triangle always pass 
through two fixed points ; prove that the third side always touches a fixed circle. 
(/3) Two sides of a given triangle touch two fixed circles ; prove that the third 
side also touches a fixed circle. 
(y) Two sides of a given polygon touch fixed circles ; prove that all the remaining 
sides also touch fixed circles.