whence

AE/AG

BE BG

or [ABEG]=X2.

But [ABEG]=P. ABEG-P. ABCD; therefore, etc.]

137. Directive Axis.-For any two homographic rows of points ABC..., A'B'C'... on different axes L and L', if any pair of corresponding points A and A' be each joined to all the points on the other axis, the two pencils A. A'B'C'..., A'. ABC... are in perspective (Art. 136), i.e. the intersections of the pairs of lines AB', A'B (C′′); AC′, A'C (B'); AD', A'D, etc., are collinear. We are thus enabled to find a point P' on the line L' corresponding to a given point P on L.

For having obtained the line B"C", join A'P and let it meet B"C" in P"; then AP" meets the axis L' in the required point.

An important point arises out of the consideration of the correspondents to the intersections O, P, and P' of the axes L, L, L" taken in pairs. By means of the general method given above we find that P on the axis L corresponds to O on the axis L', and that P' on the axis L' corresponds to O on the axis L. This shows that the axis L" of perspective of the pencils

A. A'B'C'..., A'. ABC....

whose vertices A and A' were arbitrarily chosen as any pair of correspondents of the given homographic systems, is a fixed line, since it meets each axis in a point corresponding to their intersection O regarded as a point on the other. Hence: all pairs of corresponding connectors (XY', X'Y) of pairs of non-corresponding points lie on a line. This line is called the Directive Axis of the given homographic systems.

Otherwise thus: Take the two homographic pencils at A" and L and L'as transversals to them respectively, then

[BCPO]=[C'B'P'0];

similarly for the vertex B" it follows that [CAPO]=[A'C'P'O], therefore by division (Art. 133, Ex. 3) [ABP0]=[B'A'P'O], i.e. the lines AB', A'B, PP' are concurrent.

The same proof applies to the more general case of two systems of points on a conic.

138. Directive Centre.-The following property of two homographic pencils is derived from Art. 137 by

reciprocation :-For any two homographic pencils of rays O. ABC... and O'. A'B'C'... the lines joining pairs of cor

responding intersections (AB', A′ B) of non-corresponding rays (A, B' and A', B) are concurrent.

The point of concurrence is termed the Directive Centre of the systems, and its property just stated may be proved by methods analogous to either of those given in Art. 137 for the directive axis. These are left as useful exercises for the student.

139. Problem.-To find a point X on either axis L whose correspondent on the other is at infinity (∞ ́).

Since the lines joining A, ∞' and A', X meet on the directive axis, we have the following construction :— through A draw a parallel to L', join A' to its point of intersection with the directive axis; this line meets L in the required point.

EXAMPLES.

1. Having given two homographic pencils of rays at different vertices; to find a ray of either corresponding to a given one of the other.

[By means of their directive centre.]

2. If two homographic rows of points are such that the points ∞, 'at infinity on the axis correspond, the lines are divided similarly. [For [ABC]=[A'B'C''], hence AB: BC= A'B': B'C'; therefore, etc.]

3. Having given the vertical angle in magnitude and position of a triangle of constant species, the extremities of the base divide the sides homographically.

4. If the lines AA', BB', CC" connecting the corresponding vertices of two triangles ABC and A'B'C' are concurrent at a point 0, the intersections X, Y, Z of the pairs of sides BC, B'C', etc., are collinear (cf. Art. 66).

[Join XY and let it meet the lines AA', BB', CC' in X', Y, Z respectively. Then

X.OBY'B'=X. OCZ'C' = Y. OCZ'C' Y. OAX'A';

=

therefore [OBY'B']=[OAX'A'], and since the point 0 is common to both rows the pairs of connectors AB, X'Y', A'B' are concurrent.

Therefore also the centre 0 and axis of perspective L of the two triangles divide the corresponding segments AA', BB', CC' equianharmonically.]

5. A variable triangle moves with its vertices on three concurrent lines such that two of its sides pass through fixed points X and Y; then the third side passes through a fixed point on the line XY.

6. The lines joining pairs of corresponding points of any two figures in perspective are cut homographically by the centre and axis of perspective.

7. Any line passing through either centre of perspective of two circles is cut in a constant anharmonic ratio by their radical axis. 8. Every four of the six points X, Y, Z, X', Y', Z' in Ex. 4 are equianharmonic with their four opposites.

S

9. In the figure of Art. 137 prove the relations

1°. [BCPO]=[B'C'OP'] = [B"C"P'P],

[CAPO]=[C'A'OP′] =[C"A" P'P].

2°. [ABCP]=[A'B'C'O] =[A"B"C" P],

[ABCO]=[A'B'C'P']=[A′′B"C"P'].

NOTE. It will be seen that the triangle ABC" is inscribed to A'BC and escribed to B'C'A", and more generally that of this system of three triangles each is inscribed to one and escribed to the other of the remaining two.

The vertex A and opposite side B"C" of the triangle A'B'C" form with the extremities B and C of the corresponding side of A'BC to which it is inscribed a row of points B, C, A, P. Similarly the vertex A' and opposite side BC of A'BC form with the corresponding side B'C' of the triangle A"BC" to which it is inscribed a row B', C', A', O. But these rows are equianharmonic (Ex. 8, 2°); hence for such a system of triangles the vertex and the opposite side of each divide homographically the corresponding side of the triangle to which it is inscribed.

Again, B'C"PP' is the row of points formed by the extremities of the base B"C" and its intersections with the corresponding sides BC and B'C' of the remaining triangles. But

B"C"PP-BCP0=B'C'OP ;

hence the sides of each are cut homographically by the corresponding sides of the other two.

if two of its

Let the point C' vary along the axis L'. Then the lines AC' and BC' turn around the fixed points A and B ; A" and B" move on the lines A'C and BC", and the directive axis passes through the fixed point C". In this case A'B'C' is a variable triangle inscribed to A'B'C' and escribed to ABC", both of which are fixed. Hence for a variable triangle A"B"C" inscribed to a given one A'B'C, sides pass through the vertices A and B of a triangle escribed to the latter, its third side passes through the third vertex C". Let us now consider two positions of the variable triangle A"B"C". Since its sides pass through the fixed points A, B, C" respectively, ABC" is a common inscribed triangle. Hence when two triangles are each inscribed to a third A'B'C', if the sides A"B", etc., and opposite vertices C', etc., divide the corresponding side A'B' of A'B'C' in a constant