which is impossible unless d and d, coincide.

de do

Two lines divided so that the anharmonic ratio of any four points on the one is equal to the anharmonic ratio of the four corresponding points on the other are said to be divided homographically.

PROP. 5. If the anharmonic ratios of two pencils of four rays, corresponding each to each, are equal, and the pencils are so placed that two corresponding lines coincide in direction, the three points of intersection of the remaining homologous rays lie on a straight line (Fig. 122).

1

Let 0 and 0, be the vertices of the pencils, the ray OA of the one coinciding in direction with the ray Oa of the other, and let the

1

homologous rays OB, 0,6 meet in b, and OC, 0 ̧c in c; the straight line bc will pass through the point d where OD intersects 0,d, for if not, let it meet OD in D and O̟,d in d1; then, since the anharmonic ratios of the two pencils are equal, we have

ab Db ab db

ас Dc ac d.c

which is impossible unless d, and D coincide.

1

Two pencils such that the anharmonic ratio of any four rays of the one is equal to the anharmonic ratio of the corresponding four rays of the other are said to be homographic.

218

HOMOGRAPHIC DIVISION.

PROBLEM 121. Given any number of points A, B, C, D, E... on a straight line, and any three corresponding points (as a, b, c) on a second line, to complete the homographic division of the second line (Fig. 121).

Place the lines at any angle with each other and with two corresponding points (as A and a) coinciding. Let the lines joining the remaining pairs of corresponding points (Bb and Cc) meet in O, then the lines OD, OE...&c. will meet the second line in d, e, &c., the required points of homographic division.

The construction is obvious from the known property of the transversals of a pencil of rays.

It follows conversely that if a pencil of rays intersect any two lines in points B, b; C, c...... the lines are divided homographically, and that the point A in which the two lines intersect is its own homologue in both divisions.

PROBLEM 122. Given a pencil of rays O. ABCDE... and any three corresponding rays 0 ̧.abc of a second pencil, to complete the second pencil so that the two shall be homographic (Fig. 122).

Place the pencils so that two corresponding rays (04, 0,a suppose) shall be coincident in direction, let the homologous rays intersect in b and c; the straight line bc will intersect the remaining rays of the given pencil in points on the required completing rays of the second. The construction is obvious from the known property of a transversal.

Conversely, when the corresponding rays of two pencils intersect in points on a straight line, the pencils are homographic, and the line 00, joining the centres is common to both pencils and is coincident with its own homologue.

PROP. 6. If A, B, C, D are four points on the circumference of a circle, and from any point O on the circumference the pencil O. ABCD is drawn, the anharmonic ratio of this pencil is constant.

1

For if 0, is any other point on the circumference the pencil 01. ABCD is equiangular with the pencil O. ABCD.

PROP. 7. If four fixed tangents to a circle are met by any variable tangent in A, B, C, D, the anharmonic ratio of this range is constant.

For the angles which the four points subtend at the centre are constant, and therefore the ranges are transversals of equiangular pencils.

If we reciprocate these theorems (Prob. 119), the four fixed points in the first correspond to four fixed tangents to a conic, the variable point O corresponds to a variable tangent, the lines OA, OB, &c. correspond to the points a, b, c, d in which the variable tangent cuts the fixed tangents; and since points corresponding to lines lie on lines through the centre of the auxiliary circle perpendicular to the lines to which they correspond, the pencil formed by joining a, b, c, d to the centre of the auxiliary circle is equiangular with the pencil O. ABCD, i.e. the anharmonic ratio of the range abcd is constant.

PROP. 8. The reciprocal theorem to Prop. 6 therefore is "The anharmonic ratio of the points in which four fixed tangents to a conic cut any variable tangent is constant.”

PROP. 9. By exactly similar reasoning the reciprocal theorem to Prop. 7 is "The anharmonic ratio of the pencil formed by joining any four fixed points on a conic to a variable fifth is constant."

If the reciprocal figures be drawn, by observing the angles which correspond to the constant angles in the circle, it will be seen that the angles which the four points of the variable tangent in the first theorem subtend at either focus are constant; and that the angles are constant which are subtended at the focus by the four points in which the inscribed pencil meets the directrix in the second theorem.

HOMOGRAPHIC RANGES IN THE SAME STRAIGHT LINE.

POINTS.

:

DOUBLE

When two lines divided homographically are superposed, there exist, in general, two points, each of which considered as belonging to the first division coincides with its homologue in the

second division. They may be called double points, since each represents two coincident homologous points. The double points may become imaginary.

Let A, B, C, D... (fig. 123) be any points on a straight line, S any point, and AJ a line through A making any angle with AD, and meeting SB, SC, &c. in ß, y..... respectively.

The ranges ABCD..... and Aßyd..... are of course homographie, and if the second range be rotated round A till it coincides in direction with AD, the point A, considered as belonging to the first division, will evidently coincide with its own homologue in the second division, as will also the point I determined by drawing SL perpendicular to the line bisecting the angle DAJ.

Two homographic ranges in the same straight line formed as above possess therefore two double points.

Instead however of the two ranges being formed merely by the rotation of the second about A, the second may in addition be moved along AD into any position to the right or left, bringing (as in the figure) the point corresponding to A to a, ẞ to b, y to c.... In this case also two double points in general exist, which may be thus found :

PROBLEM 123. Given two homographic ranges ABCD......., abcd... in the same straight line, to determine the double points (Fig. 124).

Draw any circle whatever, and from any point M on it draw

MA, MB, MC cutting the circumference in A,, B1, C1, and draw Ma, Mb, Mc cutting the circumference in a,, bi,

Fig. 124.

M

b1, c1

The

Draw Ab1, Ba, intersecting in K, and 4,c,, C,a, intersecting in L, and draw KL cutting the circumference in P and P1. lines MP, MP, will cut Aa in the required double points J

and J.

1

For {ABCJ}

= M{A,B,C,P} = a, {A,B,C,P}

= {NKLP},

where N is the point of intersection of KL and 1⁄4 ̧a,

and

{abcJ}=M {a,b,c,P} = A, {a,b,c,P}

= {NKLP},

(Prop. 6.)

i.e. the point J, considered as belonging to the first range, coincides with its own homologue in the second and is therefore a double point. Similarly for the point J.

It will be seen that the line PP, is a Pascal line (Prob. 86) in the circle, for C11 and B1c, intersect on it.

The double points may also be determined thus:-On Ab, Ba as chords describe two circles passing also through any arbitrary point G and intersecting again in G1, on Ac, Ca as chords describe circles passing through G and intersecting again