H being right angles, the point E must fall somewhere in GB. Again since É FI=G H B being complements of equal angles (21), the point E must fall somewhere in H G. Now since È is to be at the same time in B G and H G, it must be at their intersection G. Hence E F G H. Before leaving this proposition, we will observe that it explains the nature of an instrument, by which parailel lines are drawn with much greater facility than by the process described in the preceding article. Let A B and C D represent two parallel pieces of wood, and E F and G H two parallel cross-pieces. The cross-pieces are connected with the parallels by pivots at each of their extremities E, F, G, H. Then by varying the obliquity of the crosspieces, the distance between the two parallels may be varied at pleasure, without destroying their parallelism. 39. — Two angles which have their sides parallel and directed the same way, are equal— Let the two angles be ED F and B A C (fig. 24). Produce E D to G. Since B A F 24 and E G are parallel, the angle B AC=E G C (34). Again since D F and A C are parallel, the angle E D F=E G C. Tberefore B A C=E D F, each being equal to E GC. 40. If a line E F (fig. 25) touches the circumference F25 of a circle only in one point I, it is called a tangent. We draw a tangent to any point, by making it perpendicular to the extremity of the radius at that point. For since every line G K drawn to a point different from I, would be an oblique line, and therefore greater than the radius G I, it follows that I is the only point common to the straight line and the curve. But if a straight line cuts the circumference in two points, as C D or A B, it is then called a secant. These definitions being kept in mind, we proceed to demonstrate the following proposition--two parallels, whether tangents or secants, intercept upon the circumference equal arcs We shall first take the case of two secants. We say then that the arc L M-N 0. For bisect the chord MO by the perpendicular GH, and you bisect also its arcMHO (28). Therefore M H=H 0. Again G H is also perpendicular to L N (36), and passes through the middle of it; for if it did not, a perpendicular might be erected at the middle of L N, which would pass through the centre G (28), and then we should have two perpendiculars drawn from the same a a We say point to the same straight line, which is impossible (32). 41. —An angle which has its vertex in the circumference of a circle, has for its measure half the arc intercepted between its sides If an angle has its vertex in the circumference, it must either be formed by a tangent and a chord, as B F26 A C (fig. 26), or it must be formed by two chords, as B A I. We shall demonstrate that the proposition above enunciated is true in both cases. First we say that the angle B A C formed by a tangent and a chord, has for its measure half the arc B G A. Draw the diameter DE parallel to B A, and the diameter F G parallel to A C. The angle B A C=D HG (39). Therefore BAC has for its measure an arc equal to D G. It only remains, then, to prove that D G=half of B G A the intercepted arc. Now if D G be taken from B G A, we have remaining B D+A G. But B D+A G=F E, since B D=A E (40), A G=A F, and A E+A F=F E. Moreover F E=D G, since they measure vertical angles. Therefore B D+A G=D G, and D G is half of B G A. But D G measures BAC. Therefore B A C is measured by half of B GA, that is by half the arc comprehended between its sides. We are next to show that B A I, formed by two chords, has for its measure half of B I. Now BÀ I is the difference between I A C and B A C, and must therefore have for its measure the difference between their measures, that is, the difference between half of I G A and half of BGA, which is half of B I. 42. If an angle be formed by two chords, as A CB; F27 (fig. 27), it is called an inscribed angle. We conclude, then, as a corollary from the preceding proposition, that—all angles inscribed in the same segment, are equal--and that-all angles inscribed in a semicircumference are right an gles. Thus the angles A C B, A D B, A G B are equal, because they have for their measure half the arc A F B. Also A B F and A E F are equal, being measured by half of A F. Again, A B being a diameter, the angles A CB, A D B, A Ğ B are right angles, because their measure is half the semicircumference A F B, that is a quadrant. 43. --An angle, whose vertex is between the centre and the circumference, has for its measure half the intercepted arc, plus half the arc contained between its sides produced; and an angle whose vertex is without the circumference, has for its measure half the concave arc intercepted between its sides minus half the convex arc— First we are to prove that B AC (fig. 28) has for its measure half of B C+half of H I. F28 Produce B A and C A, and draw K H parallel to B 1. Then B A C=K H C (39). But K H C has for its measure (41) half of (K B+B C). Now since K B=H I (40), we have half of (K B+B C)=half of (H I+B C), which was to be proved. In the second place, we are to prove that C D E has for its measure half of C E-half of L F, or half of (C E-L F). Draw G F parallel to C C C D. Then C D E=G F E (39). But G F E is measured by half of G E. Now since GE=C E-C G=C E- F, half of G E=Wulf of (CE-L F), which was to be proved. Triangles. 44. The least number of straight lines that can enclose a space is three. Two make an angle or opening, and a third is necessary to close up that opening. Thus B C (fig. 29) closes up the opening made by A B and A C. F29 Such a figure is called a triangle, from its having three angles. We shall consider it in the present section merely as a figure bounded by lines, without regard to the quantity of surface it contains. If the three sides of the triangle are equal, it is called equilateral. If only two of the sides are equal, it is ealled isosceles. If no two of the sides are equal, it is called scalene. If one of the sides be produced, as A C, the angle B C D is called an exterior angle. It is important that the properties of triangles be well understood, because as we shall see hereafter, all figures bounded by straight lines may either be divided into several triangles, or reduced to one equivalent triangle. a a 45. -Every triangle may be inscribed in a circle. A triangle is said to be inscribed in a circle when it has its F 30 three vertices in the circumference as A B C (fig. 30). Now we have already shown (29) in what manner the circumference of a circle may be made to pass through any three points not in the same straight line. Therefore, since the three vertices of a triangle can never be in the same straight line, it follows that the circumference of a circle may be made to pass through them. The triangle will then be inscribed. Thus every triangle can be in<scribed in a circle. 46. - The sum of the three angles of a triangle is always F30 equal to two right angles— The triangle A B C (fig. 30) being inscribed, each of its angles is measured by half the arc contained between its sides (41). Thus A is measured by half the arc B C, B by half the arc A C, and C by half the arc A B. But these three arcs make up the whole circumference. Therefore the three angles have for their measure a semicircumference. Hence they must be equal to iwo right angles. 47. -No triangle can have mo han one right angle, This follows as a corollary from preceding. For if a triangle could have two right anes, the remaining angle would be nothing, that is, the sides could never meet. If a triangle has one right angle, it is called a right-angled triangle. But we shall use the expression riglåt triangle, being shorter than the other and equally definite. In a F34 right triangle, as B A C (fig. 34) the side B C opposite to the right angle, is called the hypothenuse. Moreover, the two acute angles B and C being together equal to a right angle, according to the preceding proposition, we say that-each of the acute angles in a right triangle is a complement of the other- 48. -If two angles of one triàngle are equal to two angles of another, the remaining angles are equal. For each is . what remains after taking equal sums from two right angles or 180°. If, having two angles of a triangle given, we wish to find the third, it may be done arithmetically, by adding the degrees in the given angles and then subtracting their sum from 180o; or it may be done geometrically, by taking a semicircumference and cutting off two arcs equal to those which are used to measure the given angles. Then the remaining arc will be the measure of the angle required. This is very readily done by means of a protractor (18) 49. — The exterior angle is equal to the sum of the two opposite interior angles. In the triangle A B C (fig. 29), F 29 B C D is the exterior angle, and A and B are the two opposite interior angles. We say that B C D=A+B. For D if B D be taken from 180°, B C A will remain. Also if A4 B be taken from 180°, B C A will remain. Now those things which, when taken from the same thing, leave equal remainders, must themselves be equal. Therefore B C D=A+B. 50. -If a triangle is isosceles, the angles opposite to the equal sides are equal— If the side A B=the side AC (fig. 30), we say the angle B=the angle C. For since F30 the chords A B and A C are equal, the arcs A B and AC are equal. Then half the arco A B=half the arc A C. But these measure the angles B and C. Therefore the angle B=the angle C. By similar reasoning we prove the converse of this proposition, namely—if two angles of a triangle are equal, the triangle is isosceles—. For if the angles B and C are equal, the arcs A B and A C are equal. Then the chords A B and A C are equal, and the triangle is isosceles. 51. -If a triangle is equilateral it is equiangular. If the three chords are equal (fig. 30) the three arcs gre equal. Then their halves, which measure the three angles must be equal. Consequently the angles themselves must be equal." Conversely--if a triangle is equiangular it is equilateral—. If the triangle A B C is equiangular, the three arcs are equal. Then the three chords must be equal, and the triangle is equilateral. 52. —In any triangle the greater side is opposite to the greater angle“. If the angle B (fig. 31) is greater than F31 A, we say that the side A C is greater than the side B C. For in this case the arc A C must be greater than the arc B C, since half of A C measures a greater angle than half of B C. But then the chord A C must be greater than the chord B C. Conversely—in any triangle the greater angle is opposite to the greater side. For if the chord A C exceeds the chord B C, the arc A C must exceed We arc B. C. Then half of A Ć exceeds half of BC. Consequently B exceeds A, which was to be proved. 53. — Two triangles are equal, when they have two sides re Fe |