cutting DC in C. Make DCA LADC, and join B C ; then A B C is the triangle required. 29. Construct a triangle, having given one angle, the perimeter, and the altitude of the triangle from one of the other angles. Make an angle, A B C, equal to the given angle. C A B C G A parallel to B C. At B raise a perpendicular, B G, and make B G equal to the given height. D Through G draw Continue B C to D, so that BD+BA the given perimeter. Join A D and make at ACAD= LADC. the triangle required. Then A B C is Proof. Because CAD ADC.. AC = C D. = = ...AB+ B C + CA= the given perimeter. Also the height of the triangle = BG the given height, and ABC the given angle. = 30. Construct a triangle, having given the sum of two sides and the angles. Take BD equal to the sum of the two given sides (see preceding fig.); at D make BDA= half the angle contained by the two sides whose sum is given; at B make DBA = one of the other angles. A B C be the triangle required. Proof. A B C = one given angle by construction. Since ACCD..BC+CA=BD = the given sum. Also BCA = ≤ CAD+ ▲ CD A = twice BDA the other given angle. 31. Construct a triangle, having given the angles and perimeter. The analysis of Problem 28 shows the construction to be as follows: Take D E the given perimeter; at D make CDA = half one of the given angles; at E make ▲ CED half another. From C draw CA and C B. making = <DCA = 4 CDA and ▲ BCE = / CEB. Then A B C shall be the triangle required. CHAPTER IX. THE CIRCLE. SECTION I.-Chords and Secants. 65. If two chords in a circle bisect each other, they both pass through the centre. Let AB and CD be chords bisecting each other in O, then shall O be the centre. If O be not the centre, let E be the centre, and join E O. E C A B D Then, since EO is part of a diameter which bisects both A B and C D, it is perpendicular to both of them, which is impossible. Therefore E is not the centre, and no point but O can be the centre. 66. If two chords, A B and C D, intersect within a circle, the sum AC + BD of the arcs which they intercept is equal to the sum of the arcs intercepted by the diameters parallel to the chords. Let MN be parallel to DC, and PQ parallel to A B.. .. Arc A C+ arc D B = arc M P + arc N Q. 67. Find the locus of the middle points of parallel chords in a circle. Let M N be the diameter perpendicular to one of the chords; then M N is perpendicular to all. First let A B be any one of the chords, and let it cut M N in C; then, by Theorem XXIII., A B is bisected in C. Conversely, let C be any point M A B C N in M N, and AB the chord which will be bisected in C; then, by Theorem XXIV., A B will be perpendicular to CD, and will therefore be one of the parallel chords. Since chords parallel to A B are bisected by M N and all chords bisected by M N are parallel to A B, F therefore M N is the locus of the middle points of all chords parallel to M N. 68. Two equal chords intersect within the circle: prove that the parts of the one are respectively equal to the parts of the other. D Let A B and C D be equal chords intersecting in E. Join A C, B D. E Because A B = CD, arc A CB= arc CA B. Take away the arc A C, .. Arc AD = arc C B ; .. ¿CDB = LABD: 69. The least chord that can be drawn through a given point is that which is perpendicular to the diameter through the given point. Let A B be perpendicular to CO, and let EF be any other chord through C, and OD the perpendicular on it from |