cases. First, however, we shall give a few theorems which relate to the general correspondence, not to the perspective position. 28. Two rows or pencils, flat or axial, which are projective to a third are projective to each other; this follows at once from the definitions. $29. If two rows, or two pencils, either flat or axial, or a row and a pencil, be projective, we may assume to any three elements in the one the three corresponding elements in the other, and then the correspondence is uniquely determined. For if in two projective rows we assume that the points A, B, C in the first correspond to the given points A', B', C'in the second, then to any fourth point D in the first will correspond a point D' in the second, so that (AB, CD) = (A'B', C'D'). But there is only one point, D', which makes the cross-ratio sponds to any given 30. If two rows are perspective, then the lines joining corre- viz. For let A. B, C, D... be points in the one, and A', B', C', D'... the corresponding points in the other row, and let A be made to coincide with its corresponding point A'. Let S be the point where the lines BB' and CC' meet, and let us join S to the point D in the first row. This line will cut the second row in a point D", so that A, B, C, D are projected from S into the points A, B, C, D. The cross-ratio (AB, CD) is therefore equal to (AB', C'D"), and by hypothesis it is equal to (A'B', C'D'). Hence (A'B', C'D")=(A'B', C'D'), that is, D is the same point as D'. $31. If two projected flat pencils in the same plane are in perspective, then the intersections of corresponding lines form a row, and the line joining the two centres as a line in the first pencil corresponds to the same line as a line in the second. And conversely, If two projective pencils in the same plane, but with different centres, have one line in the one coincident with its corresponding line in the other, then the two pencils are perspective, that is, the intersection of corresponding lines lie in a line. The proof is the same as in § 30. $32. If two projective flat pencils in the same point (pencil in space), but not in the same plane, are perspective, then the planes joining corresponding rays all pass through a line (they form an axial pencil), and the line common to the two pencils (in which their planes intersect) corresponds to itself. And conversely:If two flat pencils which have a common centre, but do not lie in a common plane, are placed so that one ray in the one coincides with its corresponding ray in the other, then they are perspective, that is, the planes joining corresponding lines all pass through a line. $33. If two projective axial pencils are perspective, then the intersection of corresponding planes lie in a plane, and the plane common to the two pencils (in which the two axes lie) corresponds to itself. And conversely: If two projective axial pencils are placed in such a position that a plane in the one coincides with its corresponding plane, then the two pencils are perspective, that is, corresponding planes meet in lines which lie in a plane. The proof again is the same as in § 30. $34. These theorems relating to perspective position become illusory if the projective rows of pencils have a common base. We then have: In two projective rows on the same line-and also in two projective and concentric flat pencils in the same plane, or in two projective axial pencils with a common axis-every element in the one coincides with its corresponding element in the other as soon as three elements in the one coincide with their corresponding elements in the other. K A C' B' ૧ FIG. 9. element in the other. S If two flat pencils, S, and S, in a plane are perspective (fig. 10), we need only to know two pairs, a, a' and b, 'b', of corresponding rays in order to find the axis s of projection. This being known, a ray c' in S, corresponding to a given s ray c in Si, is found by joining S to the point where c cuts the axis s. A similar construction holds in the other cases of perspective figures. On this depends the solution of the following general problem. § 36. Three pairs of corresponding elements in two projective rows or pencils being given, to determine for any element in one the corresponding element in the other. We solve this in the two cases of two projective rows and of two projective flat pencils in a plane. Problem I.-Let A, B, C be Problem II.-Let a, b, c be three points in a row s, A', B', C' three rays in a pencil S, a', b', c' the corresponding points in a the corresponding rays in a proprojective row s', both being in a jective pencil S', both being in plane; it is required to find for the same plane; it is required to any point D in s the correspond- find for any ray d in S the corre ing point D' in s'. sponding ray d'in S'. The solution is made to depend on the construction of an auxiliary row or pencil which is perspective to both the given ones. This is found as follows: S' Solution of Problem I.-On the line joining two corresponding points, say ÁA' (fig. 11), take any two points, S and S', as centres of auxiliary pencils. Join the intersection B of SB and S'B' to the intersection C1 of SC and S'C' by the line s Then a row on s1 will be perspective to s with S as centre of projection, and to s' with S' as centre. To find now the point D' on s' corresponding to a point D on s we have only to determine the point Di, D. where the line SD cuts Proof (in case of two rows).-Between four elements A, B, C, D and their corresponding clements A', B', C', D' exists the relations, and to draw S'D1; (ABCD) = (A'B'C'D'). If now A', B', C' coincide respectively with A, B, C, we get (AB, CD) = (AB, CD'), hence D and 'D' coincide. The last theorem may also be stated thus: In two projective rows or pencils, which have a common base but are not identical, not more than two elements in the one can coincide with their corresponding elements in the other. Thus two projective rows on the same line cannot have more than two pairs of coincident points unless every point coincides with its corresponding point. It is easy to construct two projective rows on the same line, which have two pairs of corresponding points coincident. Let the points A, B, C as points belonging to the one row correspond to A, the point where this line Proof. The rows and s' are both perspective to the row s1, hence they are projective to one another. To A, B, C, D ons correspond A, B, C, D on si, and B FIG. 11. to these correspond A', B', C', D' on s'; so that D and D' are corresponding points as required. S Solution of Problem II.-Through the intersection A of two corresponding rays a and a' (fig. 12), take two lines, s and s', as bases of auxiliary rows. Let S be the point where the line b, which joins B and B', cuts the line c, which joins C and C. Then a pencil S will be perspective to S with s as axis of projection. To find the ray d' in B a Theorem. The product of the distances of any two corresponding points in two projective rows from the points which correspond to the points at infinity in the other is constant, viz. AJ. A'l'=k. Steiner has called this number k the Power of the correspondence. [The relation AJ. A'I'=k shows that if J, I' be given then the point A' corresponding to a specified point A is readily found; hence A, A' generate homographic ranges of which I and J' correspond to the points at infinity on the ranges. If we take any two origins 0, O', on the ranges and reduce the expression AJ. A'Í'=k to its algeS' corresponding to a given ray dbraic equivalent, we derive an equation of the form axx+6x+x' in S, cut d by s at D; project +8=0. Conversely, if a relation of this nature holds, then points this point from S to D' on s' corresponding to solutions in x, x' form homographic ranges.] and join D' to S'. This will be the required ray. Proof-That the pencil S is perspective to S and also to S' follows from construction. Το the lines a, b, c, d in S, correspond the lines a, b, c, d in S and the lines a', b', c', d' in S', so that d and d' are corresponding rays. In the first solution the two centres, S, S', are any two points on a line joining any two corresponding points, so that the solution of the problem allows of a great many different constructions. But whatever construction be used, the point D', corresponding to D, must be always the same, according to the theorem in § 29. This gives rise to a number of theorems, into which, however, we shall not enter. The same remarks hold for the second problem. FIG. 12. $37. Homological Triangles.-As a further application of the theorems about perspective rows and pencils we shall prove the following important theorem. Theorem. If ABC and A'B'C' (fig. 13) be two triangles, such that the lines AA', BB', CC' meet in a point S, then the intersections of BC and B'C', of CA and C'A', and of AB and A'B' will lie in a line. Such triangles are said to be homological, or in perspective. The triangles are "co-axial" in virtue of the property that the meets of corresponding sides are collinear and copolar, since the lines joining corresponding vertices are concurrent. Proof.-Let a, b, c denote the lines AA', BB', CC', which meet at S. Then these may be taken as bases of projective rows, so that A, A', S on a correspond to B, B', S on b, and to C, C', S on c. As the point S is common to all, any two of these rows will be perspective. If S, be the centre of projection of rows b and c, c and a, and if the line SS, cuts a in A,, and b in B, and c in C, then A, B, in a and b, both corresponding will be corresponding points to C1 in c. But a and b are perspective, therefore the line AB, that is SS2, joining corresponding points must pass through the centre of projection S, of a and b. In other words, S1, S2, Sa lie in a line. This is Desargues' celebrated theorem if we state it 16 Sa thus: Theorem of Desargues.-If each of two triangles has one vertex on each of three concurrent lines, then the intersections of corresponding sides lie in a line, those sides being called corresponding which are opposite to vertices on the same line. FIG. 13. The converse theorem holds also, viz. Theorem.-If the sides of one triangle meet those of another in three points which lie in a line, then the vertices lie on three lines which meet in a point. The proof is almost the same as before. 838. Metrical Relations between Projective Rows.-Every row contains one point which is distinguished from all others, viz. the point at infinity. In two projective rows, to the point I at infinity in one corresponds a point l' in the other, and to the point J' at infinity in the second corresponds a point J in the first. The points I' and J are in general finite. If now A and B are any two points in the one, A', B' the corresponding points in the other row, then ог But. by § 17. (AB, JI) = (A'B', J'I'), AJ/JB : AI/IB=A'J'/J'B' : A'I'/I'B'. therefore the last equation changes into that is to say AI/IB A'JJ'B'=-1; AJ. A'I'=BJ. B'I', AC/CB=A'C'/C'B' or AC/A'C' = BC/B'C', that is, corresponding segments are proportional. Conversely, if corresponding segments are proportional, then to the point at infinity in one corresponds the point at infinity in the other. If we call such rows similar, we may state the result thus Theorem.-Two projective rows are similar if to the point at infinity in one corresponds the point at infinity in the other, and conversely, if two rows are similar then they are projective, and the points at infinity are corresponding points. From this the well-known propositions follow: Two lines are cut proportionally (in similar rows) by a series of parallels. The rows are perspective, with centre of projection at infinity. If two similar rows are placed parallel, then the lines joining homologous points pass through a common point. § 40. If two flat pencils be projective, then there exists in either, one single pair of lines at right angles to one another, such that the corresponding lines in the other pencil are again at right angles. To prove this, we place the pencils in perspective position (fig. 14) by making one ray coincident with its corresponding ray. Corresponding rays meet then on a line p. And now we draw the circle which has its centre O on p, and which passes through the centres S and S' of the two pencils. This circle cuts p in two points H and K. The Sand S' will be pairs of two pairs of rays, h, k, and h', k', joining these points to corresponding rays at right The construction angles. gives in general but circle, but if the line is the perpendicular bisector of SS', there exists an infinite number, and to every right angle in the one pencil corresponds a right angle in the other. §41. It has been stated in § 1 that not only points, but also planes and lines, are taken as elements out of which figures are built up. We shall now see that the construction of one figure which possesses certain properties gives rise in many cases to the construction of another figure, by replacing, according to definite rules, elements of one kind by those of another. The new figure thus obtained will then possess properties which may be stated as soon as those of the original figure are known. We obtain thus a principle, known as the principle of duality or of reciprocity, which enables us to construct to any figure not containing any measurement in its construction a reciprocal figure, as it is called, and to deduce from any theorem a reciprocal theorem, for which no further proof is needed. It is convenient to print reciprocal propositions on opposite sides of a page broken into two columns, and this plan will occasionally be adopted. We begin by repeating in this form a few of our former state For instance, if in the first figure we take a plane and three points in it, we have to take in the second figure a point and three planes through it. The three points in the first, together with the three lines joining them two and two, form a triangle; the three planes in the second and their three lines of intersection form a trihedral angle. A triangle and a trihedral angle are therefore reciprocal figures. Similarly, to any figure in a plane consisting of points and lines will correspond a figure consisting of planes and lines passing through a point S, and hence belonging to the pencil which has S as centre. The figure reciprocal to four points in space which do not lie in a plane will consist of four planes which do not meet in a point. In this case each figure forms a tetrahedron. 42. As other examples we have the following:To a row a flat pencil is reciprocal an axial pencil, a flat pencil, a field of points and lines,, a pencil of planes and lines, the space of points the space of planes. For the row consists of a line and all the points in it, reciprocal to it therefore will be a line with all planes through it, that is, an axial pencil; and so for the other cases. This correspondence of reciprocity breaks down, however, if we take figures which contain measurement in their construction. For instance, there is no figure reciprocal to two planes at right angles, because there is no segment in a row which has a magnitude as definite as a right angle. We add a few examples of reciprocal propositions which are easily proved. Theorem.-If A, B, C, D are any four points in space, and if the lines AB and CD meet, then all four points lie in a plane, hence also AC and BD, as well as AD and BC, meet. Theorem.-If of any number of whilst all do not lie in a point, then all lie in a plane. Theorem.-If a, B, Y, & are four planes in space, and if the lines aß and yo meet, then all four planes lie in a point (pencil), hence also ay and so, as well as as and By, meet. lines every one meets every other, lie in a plane, then all lie in a point (pencil). § 43. Reciprocal figures as explained lie both in space of three dimensions. If the one is confined to a plane (is formed of elements which lie in a plane), then the reciprocal figure is confined to a pencil (is formed of elements which pass through a point). But there is also a more special principle of duality, according to which figures are reciprocal which lie both in a plane or both in a pencil. In the plane we take points and lines as reciprocal elements, for they have this fundamental property in common, that two elements of one kind determine one of the other. In the pencil, on the other hand, lines and planes have to be taken as reciprocal, and here it holds again that two lines or planes determine one plane or line. Thus, to one plane figure we can construct one reciprocal figure in the plane, and to each one reciprocal figure in a pencil.. We mention a few of these. At first we explain a few names:A figure consisting of n points A figure consisting of n lines in a plane will be called an in a plane will be called an n-side. n-point. A figure consisting of n planes in a pencil will be called an n-flat. A figure consisting of n lines in a pencil will be called an n-edge. It will be understood that an n-side is different from a polygon of n sides. The latter has sides of finite length and n vertices, the former has sides all of infinite extension, and every point where two of the sides meet will be a vertex. A similar difference exists between a solid angle and an n-edge or an n-flat. We notice particularly A four-point has six sides, of which two and two are opposite, and three diagonal points, which are intersections of opposite sides. A four-flat has six edges, of which two and two are opposite, and three diagonal planes, which pass through opposite edges. A four-side has six vertices, of which two and two are opposite, and three diagonals, which join opposite vertices. A four-edge has six faces, of which two and two are opposite, and three diagonal edges, which are intersections of opposite faces. To a surface as locus of points corresponds, in the same manner, a surface as envelope of planes; and to a curve in space as locus of points corresponds a developable surface as envelope of planes. It will be seen from the above that we may, by aid of the principle of duality, construct for every figure a reciprocal figure, and that to any property of the one a reciprocal property of the other will exist, as long as we consider only properties which depend upon nothing but the positions and intersections of the different elements and not upon measurement. For such propositions it will therefore be unnecessary to prove more than one of two reciprocal theorems. FIG. 15. GENERATION OF CURVES and Cones of Second Order § 45. Conics.-If we have two projective pencils in a plane, corresponding rays will meet, and their point of intersection will constitute some locus which we have to investigate. Reciprocally, if two projective rows in a plane are given, then the lines which join corresponding points will envelope some curve. We prove first:Theorem.-If two projective flat pencils lie in a plane, but are neither in perspective nor concentric, then the locus of intersections of corresponding rays is a curve of the second order, that is, no line contains more than two points of the locus. Proof. We draw any line. This cuts each of the pencils in a row, so that we have on two Theorem.-If two projective rows lie in a plane, but are neither in perspective nor on a common base, then the envelope of lines joining corresponding points is a curve of the second class, that is, through no point pass more than two of the enveloping lines. Proof. We take any point T and join it to all points in each row. This gives two concentric pencils, which are projective because the rows are projective. If a line joining corresponding points in the two rows passes through T, it will be a line in the one pencil which coincides with its corresponding line in the other. But two projective concentric flat pencils in the same plane cannot have more than two lines of one coincident with their corresponding line in the other (834). rows, and these are projective because the pencils are projective. If corresponding rays of the two pencils meet on the line, their intersection will be a point in the one row which coincides with its corresponding point in the other. But two projective rows on the same base cannot have more than two points of one coincident with their corresponding points in the other (§ 34). It will be seen that the proofs are reciprocal, so that the one may be copied from the other by simply interchanging the words point and line, locus and envelope, row and pencil, and so on. We shall therefore in future prove seldom more than one of two reciprocal theorems, and often state one theorem only, the reader being recommended to go through the reciprocal proof by himself, and to supply the reciprocal theorems when not given. § 46. We state the theorems in the pencil reciprocal to the last, without proving them: Theorem.-If two projective flat pencils are concentric, but are neither perspective nor coplanar, then the envelope of the planes joining corresponding rays is a cone of the second class; that is, no line through the common centre contains more than two of the enveloping planes. Theorem.-If two projective axial pencils lie in the same pencil (their axes meet in a point), but are neither perspective nor co-axial, then the locus of lines joining corresponding planes is a cone of the second order; that is, no plane in the pencil contains more than two of these lines. § 47. Of theorems about cones of second order and cones of second class we shall state only very few. We point out, however, the following connexion between the curves and cones under consideration: The lines which join any point in space to the points on a curve A four-side is usually called a complete quadrilateral, and a four-of the second order form a cone point a complete quadrangle. The above notation, however, seems better adapted for the statement of reciprocal propositions. 844 If a point moves in a plane it describes a plane curve. If a plane moves in a pencil it envelopes a cone. If a line moves in a plane it envelopes a plane curve (fig. 15). If a line moves in a pencil it describes a cone. A curve thus appears as generated either by points, and then we call it a "locus," or by lines, and then we call it an "envelope." in the same manner a cone, which means here a surface, appears either as the locus of lines passing through a fixed point, the vertex of the cone, or as the envelope of planes passing through the same point. of the second order. The planes which join any point in space to the lines enveloping a curve of the second class envelope themselves a cone of the second class. Every plane section of a cone of the second order is a curve of the second order. Every plane section of a cone of the second class is a curve of the second class. By its aid, or by the principle of duality, it will be easy to obtain theorems about them from the theorems about the curves. We prove the first. A curve of the second order is generated by two projective pencils. These pencils, when joined to the point in space, give rise to two projective axial pencils, which generate the cone in question as the locus of the lines where corresponding planes meet. $48. Theorem.-The curve of second order which is generated by two projective flat pencils passes through the centres of the two pencils. Proof. If S and S' are the two pencils, then to the ray SS' or p' in the pencil S' corresponds in the pencil S a ray p, which is different from p', for the pencils are not perspective. But p and p' meet at S, so that S is a point on the curve, and similarly S'. Theorem.-The envelope of second class which is generated by two projective rows contains the bases of these rows as enveloping lines or tangents. Proof-If s and s are the two rows, then to the point ss' or P' as a point in s' corresponds in s a point P, which is not coincident with P', for the rows are not perspective. But P and P' are joined by s, so that s is one of the enveloping lines, and similarly s'. It follows that every line in one of the two pencils cuts the curve in two points, viz. once at the centre S of the pencil, and once where it cuts its corresponding ray in the other pencil. These two points, however, coincide, if the line is cut by its corresponding line at S itself. The line p in S, which corresponds to the line SS' in S', is therefore the only line through S which has but one point in common with the curve, or which cuts the curve in two coincident points. Such a line is called a tangent to the curve, touching the latter at the point S, which is called the "point of contact." In the same manner we get in the reciprocal investigation the result that through every point in one of the rows, say in s, two tangents may be drawn to the curve, the one being s, the other the line joining the point to its corresponding point in s. There is, however, one point P in s for which these two lines coincide. Such a point in one of the tangents is called the "point of contact" of the tangent. We thus get Theorem.-To the line joining the centres of the projective pencils as a line in one pencil corresponds in the other the Theorem.-To the point of intersection of the bases of two projective rows as a point in one row corresponds in the other the tangent at its centre. point of contact of its base. $49. Two projective pencils are determined if three pairs of corresponding lines are given. Hence if a, b, c are three lines in a pencil S, and a, b, c2 the corresponding lines in a projective pencil S, the correspondence and therefore the curve of the second order generated by the points of intersection of corresponding rays is determined. Of this curve we know the two centres S and S2, and the three points aja, bib, cic, hence five points in all. This and the reciprocal considerations enable us to solve the following two problems: Problem.-To construct a curve Problem.-To construct a curve of the second order, of which five of the second class, of which five points S1, S2, A, B, C are given. tangents 1, 2, a, b, c are given. In order to solve the left-hand problem, we take two of the given points, say S and S2, as centres of pencils. These we make projective by taking the rays a, b, c, which join S, to A, B, C respectively, as corresponding to the rays a, b, c, which join S, to A, B, C respectively, so that three rays meet their corresponding rays at the given points A, B, C. This determines the correspondence of the pencils which will generate a curve of the second order passing through A, B, C and through the centres S, and S2, hence through the five given points. To find more points on the curve we have to construct for any ray in S, the corresponding ray in S. This has been done in § 36. But we repeat the construction in order to deduce further properties from it. We also solve the right-hand problem. Here we select two, viz. u, u of the five given lines, u, u, a, b, c, as bases of two rows, and the points A, B, C1 where a, b, c cut u as corresponding to the points A2, B2, C2 where a, b, c cut u. We get then the following solutions of the two problems: Solution.-Through the point A draw any two lines, u, and us (fig. 16), the first u to cut the pencil S in a row ABC, the other us to cut the pencil S2 in a row ABC. These two rows will be perspective, as the point A corresponds to itself, and the centre of projection will be the point S, where the lines BB2 and CC2 meet. To find now for any ray d, in S, its corresponding ray da in S, we determine the point Di where di cuts u1, project this point from S to D, on us and join S to D. This will be the required ray d which cuts d, at some point D on the curve. Solution. In the line a take any two points S and S as centres of pencils (fig. 17), the first S (ABC) to project the row u, the other S2 (A2B2C2) to project the row 2. These two pencils will be perspective, the line SA, being the same as the corresponding line SA, and the axis of projection will be the line u, which joins the intersection B of SB, and SB, to the intersection C of SC, and SC. To find now for any point D, in u, the corresponding point D2 in us, we draw SD, and project the point D where this line cuts u from S to . This will give the required point D, and the line d joining Di to D2 will be a new tangent to the curve. 50. These constructions prove, when rightly interpreted, very important properties of the curves in question. As A is any given point on the it, we have solved the problems: Problem.-To find the second a point in which any line through known point on the curve cuts the curve. tangent which can be drawn from any point in a given tangent to the curve. second order through them; we select two of the points as centres of projective pencils, and then one such curve is determined. It will be presently shown that we get always the same curve if two other points are taken as centres of pencils, that therefore five points determine one curve of the second order, and reciprocally, that five tangents determine one curve of the second class. Six points taken at random will therefore not lie on a curve of the second order. In order that this may be the case a certain condition has to be satisfied, and this condition is easily obtained from the construction in § 49, fig. 16. If we consider the conic determined by the five points A, S1, S2, K, L, then the point D will be on the curve if, s. and only if, the points on D1, S, D2 be in a line. D FIG. 18. (Da. This may be stated differently if we take AKS DSL (figs. 16 and 18) as a hexagon inscribed in the conic, then AK and DS will be opposite sides, so will be KS and SL, as well as SD and LA. The first two meet in D1. the others in S and D, respectively. We may therefore state the required condition, together with the reciprocal one, as follows: Pascal's Theorem.-If a hexagon Brianchon's Theorem.-If a be inscribed in a curve of the hexagon be circumscribed about second order, then the intersec- a curve of the second class, then tions of opposite sides are three the lines joining opposite vertices points in a line. are three lines meeting in a point. These celebrated theorems, which are known by the names of their discoverers, are perhaps the most fruitful in the whole theory of conics. Before we go over to their applications we have to show that we obtain the same curve if we take, instead of S1, S2, any two other points on the curve as centres of projective pencils. $52. We know that the curve depends only upon the correspondence between the pencils S, and S, and not upon the special construction used for finding new points on the curve. The point A (fig. 16 or 18), through which the two auxiliary rows u, u were drawn, may therefore be changed to any other point on the curve. Let us now suppose the curve drawn, and keep the points S1, S, K, L and D, and hence also the point S fixed, whilst we move A along the curve. Then the line AL will describe a pencil about L as centre, and the point D, a row on SD perspective to the pencil L. At the same time AK describes a pencil about K and D a row perspective to it on SD. But by Pascal's theorem D, and D, will always lie in a line with S, so that the rows described by D and D2 are perspective. It follows that the pencils K and L'will themselves be projective, corresponding rays meeting on the curve. This proves that we get the same curve whatever pair of the five given points we take as centres of projective pencils. HenceOnly one curve of the second order can be drawn which passes through five given points. Only one curve of the second class can be drawn which touches five given lines. We have seen that if on a curve of the second order two points coincide at A, the line joining them becomes the tangent at A. If, therefore, a point on the curve and its tangent are given, this will be equivalent to having given two points on the curve. Similarly, if on the curve of second class a tangent and its point of contact are given, this will be equivalent to two given tangents. We may therefore extend the last theorem: Only one curve of the second order can be drawn, of which four points and the tangent at one of them, or three points and the tangents at two of them, are given. Only one curve of the second class can be drawn, of which four tangents and the point of contact at one of them, or three tangents and the points of contact at two of them, are given. $53. At the same time it has been proved: If all points on a curve of the second order be joined to any two of them, then the two pencils thus formed are projective, those rays being corresponding which meet on the curve. Hence The cross-ratio of four rays joining a point S on a curve of second order to four fixed points A, B, C, D in the curve is independent of the position of S, and is called the cross-ratio of the four points A, B, C, D. If this cross-ratio equals-1 the four points are said to be four harmonic points. All tangents to a curve of second class are cut by any two of them in projective rows, those being corresponding points which lie on the same tangent. Hence The cross-ratio of the four points in which any tangent u is cut by four fixed tangents a, b, c, d is independent of the position of u, and is called the cross-ratio of the four tangents a, b, c, d. If this cross-ratio equals -1 the four tangents are said to be four harmonic tangents. We have seen that a curve of second order, as generated by projective pencils, has at the centre of each pencil one tangent; and further, that any point on the curve may be taken as centre of such pencil. Hence A curve of second order has at every point one tangent. A curve of second class has on every tangent a point of contact. $54. We return to Pascal's and Brianchon's theorems and their applications, and shall, as before, state the results both for curves of the second order and curves of the second class, but prove them only for the former. Pascal's theorem may be used when five points are given to find more points on the curve, viz. it enables us to find the point where any line through one of the given points cuts the curve again. It is convenient, in making use of Pascal's theorem, to number the points, to indicate the order in which they are to be taken in forming a hexagon, which, by the way, may be done in 60 different ways. It will be seen that I 2 (leaving out 3) 4 5 are opposite sides, so are 2 3 and (leaving out 4) 5 6, and also 3 4 and (leaving out 5) 6 1. If the points 1 2 3 4 5 are given, and we want a 6th point on a line drawn through 1, we know all the sides of the hexagon with the exception of 5 6, and this is found by Pascal's theorem. If this line should happen to pass through 1, then 6 and 1 coincide, or the line 6 1 is the tangent at 1. And always if two consecutive vertices of the hexagon approach nearer and nearer, then the side joining them will ultimately become a tangent. We may therefore consider a pentagon inscribed in a curve of second order and the tangent at one of its vertices as a hexagon, and thus get the theorem: $55. Of these theorems, those about the quadrilateral give rise to a number of others. Four points A, B, C, D may in three different ways be formed into a quadrilateral, for we may take them in the order ABCD, or ACBD, or ACDB, so that either of the points B, C, D may be taken as the vertex opposite to A. Accordingly we may apply the theorem in three different ways. Let A, B, C, D be four points on a curve of second order (fig. 21), and let us take them as forming a quadrilateral by taking the points in the order ABCD, so that A, C and also B, D are pairs of opposite vertices. Then P, Q will be the points where opposite sides ineet, |