The condition that these should meet in a point is found by eliminating ẞ between the first two, aud examining whether the resulting equation coincides with the third. It is cos (aẞ') cos (By') cos (ya') = cos (a'ẞ) cos (ẞ'y) cos (y'a). But the symmetry of this equation shews that this is also the condition that the perpendiculars from the vertices of the second triangle on the sides of the first should meet in a point. 55. The lines a-kß=0, and ka-B=0, are plainly such that one makes the same angle with the line a which the other makes with the line 8, and are therefore equally inclined to the bisector a- ß. Ex. If through the vertices of a triangle there be drawn any three lines meeting in a point, the three lines drawn through the same angles, equally inclined to the bisectors of the angles, will also meet in a point. Let the sides of the triangle be a, ẞ, y, and let the equations of the first three lines be la - mẞ = 0, mß — ny = 0, ny — la = 0, which, by the principle of Art. 41, are the equations of three lines meeting in a point, and which obviously pass through the points aß, By, and ya. Now, from this Article, the equations of the second three lines will be which (by Art. 41) must also meet in a point. 56. The reader is probably already acquainted with the following fundamental geometrical theorem :-" If a pencil of four right lines meeting in a point O be intersected by any transverse right line in the four points A, P, P', B, then the ratio AP.PB AP'.PB is constant, no matter how B P' P A the transverse line be drawn.” This ratio is p3.AP.PB=0A. OP. OP'. OB. sin AOP.sin P'OB; AP.PB sin A OP. sin P'OB but the latter is a constant quantity, independent of the position of the transverse line. 57. If a-kẞ=0, a-k'B=0, be the equations of two lines, will be the anharmonic ratio of the pencil formed by the four lines a, ß, a – kß, a— k'ß, for (Art. 54) but this is the anharmonic ratio of the pencil. The pencil is a harmonic pencil when = 1, for then the angle AOB is divided internally and externally into parts whose sines are in the same ratio. Hence we have the important theorem, two lines whose equations are a-kß=0, a+kß=0, form with a, ẞ a harmonic pencil. cut by any parallel to ẞ in the four points K, L, M, N, and the A В these four points, the perpendiculars from these points on a are (by virtue of the equations of the lines) proportional to k, l, m, n; and AK, AL, AM, AN are evidently proportional to these perpendiculars; hence NL is proportional to n-l; MK to m-k; NM to n - m; and LK to l-k. 59. The theorems of the last two articles are true of lines represented in the form P-kP', P-IP', &c., where P, P' denote ax+by+c, ax+by+c', &c. For we can bring P to the form x cos ay sina -p by dividing by a certain factor. The equations therefore P-kP=0, P-IP' = 0, &c., are equivalent to equations of the form a- kpß= 0, a − lpß = 0, &c., where p is the ratio of the factors by which P and P' must be divided in order to bring them to the forms a, B. But the expressions for anharmonic ratio are unaltered when we substitute for k, l, m, n; kp, lp, mp, np. It is worthy of remark, that since the expressions for anharmonic ratio only involve the coefficients k, l, m, n, it follows that if we have à system of any number of lines passing through a point, P-kP, P-IP', &c.; and a second system of lines passing through another point, Q−kQ', Q−1Q', &c., the line P-kP being said to correspond to the line Q-kQ', &c.; then the anharmonic ratio of any four lines of the one system is equal to that of the four corresponding lines of the other system. We shall hereafter often have occasion to speak of such systems of lines, which are called homographic systems. 60. Given three lines a, B, Y, forming a triangle; the equation of any right line, ax+by+c=0, can be thrown into the form la + mB + ny = 0. Write at full length for a, B, y the quantities which they represent, and la+mẞ+ny becomes (l cosa +m cosẞ+n cosy) x + (l sina + m sinß + n siny) y − (lp +mp' + np") = 0. This will be identical with the equation of the given line, if we have I cosa+m cosẞ+n cosy = a, l sina + m sinß + n siny=b, and we can evidently determine l, m, n, so as to satisfy these three equations. The following examples will illustrate the principle that it is possible to express the equations of all the lines of any figure in terms of any three, a=0, B = 0, y = 0. Ex. 1. To deduce analytically the harmonic properties of a complete quadrilateral. (See figure, next page). Let the equation of AC be a = 0; la mẞ0; and of BC, mẞny = 0. these quantities the equations of all the other lines of the figure. of AB, ẞ=0; of BD, y=0; of AD Then we are able to express in terms of We say "forming a triangle," for if the lines a ẞ, y meet in a point, la + mß+ny must always denote a line passing through the same point, since any values of the coordinates which make a, ß, y separately = 0, must make la + mẞ + ny = 0. For instance, the equation of CD is la - mẞ + ny = 0, for it is the equation of a right line passing through the intersection of la – mẞ and y, that is, the point D, and of a and mẞ - ny, that is, the point C. Again, la – ny = 0 is the equation of OE, for it passes through ay or E, and it also passes through the intersection of AÐ and BC, since it is = (la mẞ) + (mß — ny). EF joins the point ay to the point A B (la - mẞ+ny, ẞ), and its equation will be found to be la + ny = 0. From Art. 57 it appears that the four lines EA, EO, EB, and EF form a harmonic pencil, for their equations have been shown to be is a = 0, y = 0, and la±ny = 0. Again, the equation of FO, which joins the points (la+ny, ß) and (la—mß, mß—ny) la - 2mẞ + ny = 0. Hence (Art. 57) the four lines FE, FC, FO, and FB are a harmonic pencil, for their equations are la mẞ + ny = 0, B = 0, and la − mß + ny ± mß = 0. Again, OC, OE, OD, OF are a harmonic pencil, for their equations are la - mẞ = 0, mẞ — ny = 0, and la – mß ± (mß – ny) = 0. Ex. 2. To discuss the properties of the system of lines formed by drawing through the angles of a triangle three lines meeting in a point. Let the equation of AB be y = 0; of AC, ẞ=0; of BC, a = 0; and let the lines OA, OB, OC, meeting in a point, M L Now we can prove that the three points L, M, N are all in one right line, who equation is la + mẞ + ny = 0, for this line passes through the points (la + mẞ — ny, y) or N; (la - mẞ + ny, or M; and (mẞ + ny — la, a) or L. The equation of CN is la + mẞ = 0, for this is evidently a line through (a, ẞ) or C, and it also passes through N, !la + mẞ + ny) — ny. Hence BN is cut harmonically, for the equations of the four lines CN, CA · CF, CB are The equations of this example can be applied to many particular cases of frequent Occurrence. Thus (see Ex. 3, p. 54) the equation of the line joining the feet of two perpendiculars of a triangle is a cos A + B cos B - y cos C = 0; while a cos A+ ẞ cos B+ y cos C passes through the intersections with the opposite sides of the triangle, of the lines joining the feet of the perpendiculars. In like manner a sin A+ẞ sin B- y sin C represents the line joining the middle points of two sides, &c. Ex. 3. Two triangles are said to be homologous, when the intersections of the corresponding sides lie on the same right line called the axis of homology; prove that the lines joining the corresponding vertices meet in a point [called the centre of homology]. Let the sides of the first triangle be a, ß, y; and let the line on which the corresponding sides meet be la + mẞ+ny; then the equation of a line through the intersection of this with a must be of the form l'a + mß + ny = 0, and similarly those of the other two sides of the second triangle are la + m'ẞ + ny = 0, la + mẞ + n'y = 0. But subtracting successively each of the last three equations from another, we get for the equations of the lines joining corresponding vertices (1 − l') a = (m — m′) ß, (m — m'′) ẞ = (n − n') y, (n — n') y = (1 — 1') a, which obviously meet in a point. 61. To find the condition that two lines la + mB + ny, l'a+m'ß + n'y may be mutually perpendicular. Write the equations at full length as in Art. 60, and apply the criterion of Art. 25, Cor. 2 (AA' + BB' = 0), when we find ll' + mm' + nn' + (mn' + m'n) cos (B − y) + (nl' + n'l) cos (y − a) +(lm' + l'm) cos (a — B) = 0. Now since B and y are the angles made with the axis of x by the perpendiculars on the lines ẞ, y, B-y is the angle between those perpendiculars, which again is equal or supplemental to the angle between the lines themselves. If we suppose the origin to be within the triangle, and A, B, C to be the angles of the triangle, By is the supplement of A. The condition for perpendicularity therefore is ll'+mm'+nn'-(mn'+m'n) cosA - (nl'+n'l) cos B-(lm'+l'm) cos C=0. As a particular case of the above, the condition that la + mß +ny may be perpendicular to y is n = m cos A+ 1 cos B. In like manner we find the length of the perpendicular from x'y |