168 Mr. W. Rutherford’s Demonstration of Pascal’s Theorem 
or rather, the products of the numerators, since the products 
of the denominators are evidently equal; for in that case the 
lines will make equal angles with the axis of z. 
Performing the operations indicated by the signs with each 
of these coefficients, we get 
a = (2ae—bd) (y, xs—2#3 y3) + (4 af—d*) (ys—ys) + (2b f—de) (#,—a,)... (9.) 
B' = (2cd—be) (a, ys—ae ys) + (4 cf—e®) (7, — a6) + (2 bf —de) (y,—y6).-.(10.) 
a! = (2ae—bad) (y, we—ai ys) + (4 af—#) (iyo) + (2 Bf —de) (a1 —a¥5)...(11.) 
B = 2cd—be) (#3 ys—Yats) + (4ef—e*) (ws—as) + (2 bf—de) (y3—ys)...(12.) 
a, 6 being the coefficients of y and 2 in (7.), without the de- 
nominators, a', 6! those of y and z in (8.). 
Again, since the lines (2, 3.) meet in the axis of x; put y = 0 
in each of these, and equate the resulting values of «, thence 
we have the relation, 
(2ed—be) (y2x3—y3 #2) + (4e f—e*) (@3—ay) + (2 bf—e d) (ys—y2:) = 0... (13.) 
Reasoning in a similar way with the lines (5, 6.), (1, 2.), and 
(4, 5.), we get the following : 
(2c d—be) (#6 ys—#3y5) + (4 ef —e*) e—a5) + (2 bf—e d) (ys—ys)= 0 ...(14.) 
(2ae—b d) (y, x2—2, y2) + (4a f—@) (y,—yo) +2 bf —ed) (w,—2xz) = 0 ...(15.) 
2ae—bd) (ysyx5—wy y5) + 4af—ad) (ys—y;,) + (2b f—e d) (ay—a,) = 0...(16.) 
If, now, we substitute the values of (2cd—Jbe), (2ae—bd) 
(4 cf—e?), and (4af—d?) derived from (13, 14, 15, 16.) in 
(9, 10, 11, 12.), and multiply out, we shall get the identity 
ap = ' B, 
hence AO, DO are in the same straight line. This esta- 
blishes completely the theorem in question. 
XXVI. Demonstration of Pascal’s Theorem relative to the 
Hexagon inscribed in a Conic Section. By Wi uiam 
RutHeRFoRD, Lsg., F.R.A.S., Royal Military Academy*. 
- the three pairs of opposite sides of a hexagon inscribed in 
a conic section be produced to meet, the three points of 
intersection will range in the same straight line. 
Let ABCD EF be a hexagon inscribed in a conic sec- 
tion, and let the sides A B, D E meet in G, the sides BC, 
E F in H, and the sides CD, A Fin K. ‘Take H for origin, 
H B, H F for the axes of positive coordinates of x and y re- 
spectively, and designate the coordinates of the six angular 
points C, B, E, F, A, D by «0, «,, 08,0 By, 2 Yyy %o Yo Te- 
spectively. Then the equations of A B, CD, DE, F A are 
* Communicated by T. S. Davies, Esq., Royal Military Academy. On 
the subject of this Paper, see Phil. Mag.S, 3. vol. xxi. p. 37; and pres. vol. 
p- 31, 
