anharmonic ratio [A'B'C'P'], the intersections of their corresponding sides determine a common inscribed triangle ABC" which is escribed to A'B'C.

And the vertex C" and opposite side AB cut the corresponding sides B"C", etc., in the above constant anharmonic ratio.

140. Theorem. For any two homographic rows of points ABC... X and A'B'C'... X', if X and X' be the points whose correspondents' and ∞ are at infinity; to prove the relations

AX. A'X'=BX. B'X' = CX. C'X' = etc.

Since A, A'; B, B′; X, ∞'; ∞∞, X' are four pairs of corresponding points [ABX∞]=[A'B'∞'X']. Expanding and reducing, this relation becomes AX/BX=1÷A'X'/B'X'; therefore AX. A'X'=BX. B'X', etc., etc.; or:-If variable points A and A' be taken on fixed lines L and L' respectively such that the rectangle under the distances from two fixed points X and X' on the lines is constant, they describe homographic systems.

COR. 1. When the vertical angle of a triangle of constant area is given in magnitude and position, the extremities of the base divide the sides homographically.

In this case the points X and X', whose correspondents are 'and, are supposed to coincide at the intersection. of the axes.

By Art. 81, Ex. 3, we see that the envelope of the base is a conic; and by Ex. 29 of the same article the curve is a hyperbola whose asymptotes are the given axes.

COR. 2. Any two homographic rows of points may be so placed that the corresponding segments A A', BB', etc., may have a common segment of harmonic section.

Place the systems so that the axes L and L' and the

points X and X' are coincident. The equations of the article are then written

XA.XA=XB.XB=XC.XC =±p.

Describe a circle with X as centre having A, A'; B, B'; etc., pairs of inverse points, and let it cut the axis in M and N. MN is the common segment of harmonic section by Art. 70, but it is imaginary when A and A' lie in opposite directions from X.

Def. Two homographic systems of points on any axis which have a common segment of harmonic section are said to be in Involution, and the corresponding points A, A'; B, B'; etc., are Conjugate Points of the Involution. We have seen in Cor. 2 that there always exists a pair of points, real or imaginary, each of which regarded as belonging to either system is coincident with its correspondent of the other. These are the Double Points (M, N) of the involution, and are connected with the systems by the equations

[MNBC]=[MNB'C'], [MNCD]=[MNC'D'], etc., etc., [MABC...]=[MA'B'C'...] and [NABC...]=[NA'B'C'...]. See Art. 133, Ex. 7.

COR. 3. In any two homographic rows of points on a common axis the double points M and N are found from the equations

*

XA. X'A' XB. X'B'... = XM. X'M=XN. X'N ;

they are therefore equidistant from X and X'.

*If the distances OA, OA' from any point O on the axis be x, x', it follows that (x - OX)(x' – OX') = const., a result of the form

Axx' +Вx+Cx' + D=0 (cf. Art. 143).

141. For any two homographic rows of points we have seen how to find the correspondent P' of any point P, a°, by means of the directive axis, Art. 137, and 8° by the formula XP. X'P'= const. It will now be proved that two given homographic rows can be generated by the revolution of either of two determinate angles around fixed vertices, the positions of the latter and the magnitude of the angles depending on the equal values [ABCD...] and [A'B'C'D'...] and the positions of the axes.

142. Problem.-If ABC... and A'B'C'... be any two homographic rows of points; to find two points such that the angles subtended at them by the segments AA', BB', etc., joining pairs of corresponding points are equal.

on L coincides with X, Similarly if Q and X' Hence the lines EX and

Let E and F be the required points; X, X' the correspondents of ∞o' and ∞ (Art. 139). Since AEA' is a constant angle, if any point P EP' is parallel to the axis L. coincide, EQ is parallel to L. EX' are equally inclined to L and L', or the angles AXE and A'X'E are equal.

Again, the angles subtended at E by any two points A and X and their correspondents A' and ' are equal (hyp.); therefore in the two triangles AEX and EA'X' we also have the angles AEX and EA'X' equal, and the triangles are similar. Hence (Euc. VI. 4)

AX/XE=EX'|X'A'

and EX. EX'=AX. A'X' const. (Art. 140).

=

Now in the triangle XEX' we are given the base XX' fixed, the difference of base angles and rectangle under the sides; therefore the vertex E is one or other of two fixed points E or F, which are obviously the opposite vertices of a parallelogram with XX' as diagonal.

COR. 1. The angles AEA', AXF, and A'X'F' are equal. For if A' and X' coincide, EA is parallel to L; therefore AEA' is equal to the angle between EX' and L or between FX and L, since EX' and FX are parallel.

COR. 2. The triangles AEA', AXF, and EX'A' are similar.

[For by similar triangles AEX and EA'X' we have AX/AE EX EA', but EX'= FX, hence

AX/AE=FX/EA',

or by alternation AX/XF=AE/EA'; therefore, etc. (Euc. VI. 6).]

COR. 3. If O denote the point of intersection of the axes L and L', the points E and F are isogonal conjugates with respect to the variable triangle OAA'.

[By Cor. 2, FAX=EAA′ and FA'X' = EA'A; therefore, etc.]

*

COR. 4. The product of the perpendiculars p and p' from E and F on the variable line AA' is constant

COR. 5. The locus of the intersection of every two rectangular positions of AA' is a circle the square of whose radius (p) is given by the equation p2= 2k2+82, where 28 EF.

=

COR. 6. A variable line cutting two fixed lines homographically cuts all positions of itself in a system of points A"B"C"... such that

[ABCD...]=[A'B'C'D'...]=[A′′B"C"D"...].

Draw the directive axis XYZ... of the system as in figure. Then OX and OA", divide the angle LOL' of the quadrilateral PXP'O harmonically (Art. 68). Similarly for OY and OB".... Hence we have [0. XY...]=[0. A"B"...] Art. 133, Ex. 7. But

[0. XY...]=[P. XY...]=[P. A'B'...].

Therefore [A'B'C'...]=[A"B"C"...].

* These properties respectively may be otherwise stated:-A variable line AA' cutting two fixed axes homographically envelopes a conic of which E and F are the foci. The locus of intersection of rectangular tangents is a circle (the Director Circle).