APPENDIX B. DEMONSTRATIONS OF PROBLEMS, ETC. MANY of the problems, &c., contained in the preceding pages, require Demonstrations. These will be given here, and will be designated by the same numbers as those of the Articles to which they refer. As many of these Demonstrations involve the beautiful Theory of Transversals, &c., which has not yet found its way into our Geometries, a condensed summary of its principal Theorems will first be given. TRANSVERSALS. THEOREM I.—If a straight line be drawn so as to cut any two sides of a triangle, and the third side prolonged, thus dividing them into six parts (the prolonged side and its prolongation being two of the parts), then will the product of any three of those parts, whose extremities are not contiguous, equal the product of the other three parts. That is, in Fig. 403, ABC being the triangle, and DF the Transversal, BEXADXCF EAXDCXBF. Fig. 403. To prove this, from B draw BG, parallel to CA. From the similar triangles BEG and AED, we have BG: BE:: AD : AE. From the similar triangles BFG and CFD, we have CD: CF :: BG: BF. Multiplying these proportions together, we have BGXCD: BEXCF::ADXBG: AEXBF. Multiplying extremes and means, and suppressing the common factor BG, we have BEX ADXCF=EAXDCXBF. These six parts are sometimes said to be in involution. If the Transversal passes entirely out side of the triangle, and cuts the prolongations of all three sides, as in Fig. 404, the theorem still holds good. The same demonstration applies without any change.* B E F B Fig. 404. A F D G E C THEOREM II.-Conversely: If three points be taken on two sides of a triangle, and on the third side prolonged, or on the prolongations of the three sides, dividing them into six parts, such that the product of three non-consecutive parts equals the product of the other three parts; then will these three points lie in the same straight line. This Theorem is proved by a Reductio ad absurdum. * This Theorem may be extended to polygons. THEOREM III.-If from the summits of a triangle, lines be drawn, to a point situated either within or without the triangle, and prolonged to meet the sides of the triangle, or their prolongations, thus dividing them into six parts; then will the product of any three non-consecutive parts be equal to the product of the other three parts. That is, in Fig. 405, or Fig. 406, AE X BF X CD=EB X FC X DA. For, the triangle ABF being cut by the transversal EC, gives the relation (Theorem I.), AE X BC X FP= EB X FC X PA. The triangle ACF, being cut by the transversal DB, gives DC X FB X PA=AD X CB × FP. Multiplying these equations together, and suppressing the common factors E B Fig. 406. A PA, CB, and FP, we have AE × BF × CD — EB × FC × DA. THEOREM IV.-Conversely: If three points are situated on the three sides of a triangle, or on their prolongations (either one, or three, of these points being on the sides), so that they divide these lines in such a way that the product of any three non-consecutive parts equals the product of the other three parts, then will lines drawn from these points to the opposite angles meet in the same point. This Theorem can be demonstrated by a Reductio ad absurdum. COROLLARIES OF THE PRECEDING THEOREMS. COR. 1.—The MEDIANS of a triangle (i. e., the lines drawn from its summits to the middles of the opposite sides) meet in the same point. For, supposing, in Fig. 405, the points D, E, and F to be the middles of the sides, the products of the non-consecutive parts will be equal, i. e., AE × BF ×.CD= DA×EB×FC; singe AE EB; BF=FC, CD=DA. Then Theorem IV. applies. COR. 2.-The BISSECTRICES of a triangle (i. e., the lines bisecting its angles) meet in the same point. For, in Fig. 405 supposing the lines AF, BD, CE to be Bissectrices, we have (Legendre IV. 17): Multiplying these equations together, and omitting the common factors, we have BF XCD X AE = FC × DA X EB. Then Theorem IV. applies. COR. 3.-The ALTITUDES of a triangle (i. e., the lines drawn from its summits perpendicular to the opposite sides) meet in the same point. For, in Fig. 405, supposing the lines AF, BD, and CE, to be Altitudes, we have three pairs of similar triangles, BCD and FCA, CAE and DAB, ABF and EBC, by comparing which we obtain relations from which it is easy to deduce BF XCD×AE =EB×FC×DA; and then Theorem IV. again applies. COR. 4.-If, in Fig. 405, or Fig. 406, the point F be taken in the middle of BC, then will the line ED be parallel to BC. For, since BFFC, the equation of Theorem III. reduces to AEXCD=EB×DA; whence AE: EB:: AD: DC; consequently ED is parallel to BC. COR. 5.-Conversely: If ED be parallel to BC, then is BF = FC. For, since AE: EB:: AD: DC, we have AE X DC=EB X AD; whence, in the equation of Theorem III., we must have BF FC. COR. 6.—From the preceding Corollary, we derive the following: If two sides of a triangle are divided proportionally, starting from the same summit, as A, and lines are drawn from the extremities of the third side to the points of division, the intersections of the corresponding lines will all lie in the same straight line joining the summit A, and the middle of the base. COR. 7.-A particular case of the preceding corollary is this: Fig. 407. A B F In any trapezoid, the straight line which joins the intersection of the diagonals and the point of meeting of the non-parallel sides produced, passes through the middle of the two parallel bases. COR. 8.-If the three lines drawn through the corresponding summits of two triangles cut each other in the same point, then the three points in which the corresponding sides, produced if necessary, will meet, are situated in the same straight line. This corollary may be otherwise enunciated, thus: If two triangles have their summits situated, two and two, on three lines which meet in the same point, then, &c. This is proved by obtaining by Theorem I. three equations, which, being multiplied together, and the six common factors cancelled, give an equation to which Theorem II. applies. Triangles thus situated are called homologic; the common point of meeting of the lines passing through their summits is called the centre of homology; and the line on which the sides meet, the axis of homology. HARMONIC DIVISION. A Fig. 408. C B D DEFINITIONS.-A straight line, AB, is said to be harmonically divided at the points C and D, Hwhen these points determine two additive segments, AC, BC, and two subtractive segments, AD, BD, proportional to one another; so that AC: BC:: AD: BD. It will be seen that AC must be more than BC, since AD is more than BD.* This relation may be otherwise expressed, thus: the product of the whole line by the middle part equals the product of the extreme parts. Reciprocally, the line DC is harmonically divided at the points B and A; since the preceding proportion may be written DB: CB :: DA : CA. The four points, A, B, C, D, are called harmonics. The points C and D are called harmonic conjugates. So are the points A and B. When a straight line, as AB, is divided harmonically, its half is a mean proportional between the distance from the middle of the line to the two points, C and D, which divide it harmonically. If, from any point, O, lines be drawn so as to divide a line harmonically, these lines are called an harmonic pencil. The four lines which compose it, OA, OC, OB, OD, in the figure, are called its radii, and the pairs which pass through the conjugate points are called conjugate radii. A Fig. 409. THEOREM V.—In any harmonic pencil, a line drawn parallel to any one of the radii, is divided by the three other radii into two equal parts. But, by hypothesis, AC: BC:: AD: BD. If four lines radiating from a point are such that a line drawn parallel to one of them is divided into two equal parts by the other three, the four lines form an harmonic pencil. * Three numbers, m, n, p, arranged in decreasing order of size, form an harmonic proportion, when the difference of the first and the second is to the difference of the second and the third, as the first is to the third. Such are the numbers 6, 4, and 3; or 6, 3, and 2; or 15, 12, and 10; &c. So, in Fig. 408, are the lines AD, AB, and AC, which thus give BD: CB:: AD: AC; or AC: CB :: AD: BD. The series of fractions, 1, 1, 3, 4, 1, &c., is called an harmonic progression, because any consecutive three of its terms form an harmonic proportion. THEOREM VI.—If any transversal to an harmonic pencil be drawn, it will be divided harmonically. Let LM be the transversal. Through K, where LM intersects OB, draw EF parallel to OA. It is bisected at K by the preceding theorem; and the similar triangles, FMK and LMO, EKN and LNO, give the proportions LM: KM::OL: FK, and LN: NK::OL:EK; whence, COROLLARY.-The two sides of any angle, together with the bissectrices of the angle and of its supplement, form an harmonic pencil. THEOREM VII.-If, from the summits of any triangle, ABC, through any point, P, there be drawn the transversals AD, BE, CF, and the transversal ED be drawn to meet AB prolonged, in F', the points F and F' will divide the base AB harmonically. A Fig. 411. E P This may be otherwise expressed, thus: The line, CP, which joins the intersection of the diagonals of any quadrilateral, ABDE, with the point of meeting, C, of two opposite sides prolonged, cuts the side AB in a point F, which is the harmonic conjugate of the point of meeting, F", of the other two sides, ED and AB, prolonged. For, by Theorem I., AF' X BD X CE = F'B X DC X EA; and THE COMPLETE QUADRILATERAL. A Complete Quadrilateral is formed by drawing any four straight lines, so that each of them shall cut each of the other three, so as to give six different points of intersection. It is so called because in the figure thus formed are found three quadrilaterals; viz., in Fig. 412, ABCD, a common convex quadrilateral; EAFC, a uni-concave quadrilateral; and EBAFD, a bi-concave quadrilateral, composed of two opposite triangles. The complete quadrilateral, AEBCDF, has three diagonals; viz., two interior, AC, BD; and one exterior, EF. F Fig. 412. E E B IN C THEOREM VIII.—In every COMPLETE QUADRILATERAL the middle points of its three diagonals lie in the same straight line. AEBCDF is the quadrilateral, and LMN the middle points of its three diagonals. From A and D draw parallels to BC, and from B and C draw parallels to |