For, let the distances AB, BC be bisected by the perpendiculars DF, EF, which must meet in some point F; for if they were parallel, the lines DB, CB, perpendicular to them would also be parallel (Cor. 2. 29. 1.), or else form but one straight line: but they meet in B, and ABC is not a straight line by hypothesis. Let then, FA, FB, and FC be drawn; then, because FA, FB meet AB at equal distances from the perpendicular, they are equal. For similar reasons FB, FC, are equal; hence the points A, B, C, are all equally distant from the point F, and consequently lie in the circumference of the circle, whose centre is F, and radius FA. It is obvious, that besides this, no other circumference can pass through the same points; for the centre, lying in the perpen B E C F dicular DF bisecting the chord AB, and at the same time in the perpendicular EF bisecting the chord BC (Cor. 1. 3. 3.), must be at the intersection of these perpendiculars; so that, as there is but one centre, there can be but one circumference. PROP. C. THEOR. If two circles cut each other, the line which passes through their centres will be perpendicular to the chord which joins the points of intersection, and will divide it into two equal parts. Let CD be the line which passes through the centres of two circles cutting each other, it will be perpendicular to the chord AB, and will divide it into two equal parts. For the line AB, which joins the points of intersection, is a chord com mon to the two circles. And if a perpendicular be erected from the middle of this chord, it will pass (Cor. 1. 3. 3.) through each of the two centres C and D. But no more than one straight line can be drawn through two points; hence, the straight line which passes through the centres will bisect the chord at right angles. COR. Hence, the line joining the intersections of the circumferences of two circles, will be perpendicular to the line which joins their centres. SCHOLIUM. 1. If two circles cut each other, the distance between their centres will be less than the sum of their radii, and the greater radius will be also less For, than the sum of the smaller and the distance between the centres. CD is less (20. 1.) than CA+AD, and for the same reason, ADAC+ CD. 2. And, conversely, if the distance between the centres of two circles be less than the sum of their radii, the greater radius being at the same time less than the sum of the smaller and the distance between the centres, the two circles will cut each other. For, to make an intersection possible, the triangle CAD must be possible. Hence, not only must we have CD<AC+AD, but also the greater radius AD<AC+CD; And whenever the triangle CAD can be constructed, it is plain that the circles described from the centres C and D, will cut each other in A and B. COR. 1. Hence, if the distance between the centres of two circles be greater than the sum of their radii, the two circles will not intersect each other. COR. 2. Hence, also, if the distance between the centres be less thar the difference of the radii, the two circles will not cut each other. For, AC+CD>AD; therefore, CD>AD-AC; that is, any side of a triangle exceeds the difference between the other two. Hence, the tri angle is impossible when the distance between the centres is less than the difference of the radii; and consequently the two circles cannot cut each other. PROP. D. THEOR. In the same circle, equal angles at the centre are subtended by equal arcs; and, conversely, equal arcs subtend equal angles at the centre. Let C be the centre of a circle, and let the angle ACD be equal to the angle BCD; then the arcs AFD, DGB, subtending these angles, are equal. Join AD, DB; then the triangles ACD, BCD, having two sides and the included angle in the one, equal to two sides and the included angle in the other, are equal: so that, if ACD be applied to BCD, there shall be an entire coincidence, the point A coinciding with B, and D common to both arcs; the two extremities, therefore, of the arc AFD, thus coinciding with those of the arc BGD, all the intermediate parts must coincide, inasmuch as they are all equally distant from the centre. A H E B F G D Conversely. Let the arc AFD be equal to the arc BGD; then the angle ACD is equal to the angle BCD. For, if the arc AFD be applied to the arc BGD, they would coincide; so that the extremities AD of the chord AD, would coincide with those of the chord BD; these chords are therefore equal: hence, the angle ACD is equal to the angle BCD (8. 1.). COR. 1. It follows, moreover, that equal angles at the centre are sub tended by equal chords: and, conversely, equal chords subtend equal angles at the centre. Cor. 2. It is also evident, that equal chords subtend equal arcs: and, conversely, equal arcs are subtended by equal chords. COR. 3. If the angle at the centre of a circle be bisected, both the arc and the chord which it subtends shall also be bisected. COR. 4. It follows, likewise, that a perpendicular through the middle of the chord, bisects the angle at the centre, and passes through the middle of the arc subtended by that chord. SCHOLIUM. The centre C, the middle point E of the chord AB, and the middle point D of the arc subtended by this chord, are three points situated in the same line perpendicular to the chord. But two points are sufficient to determine the position of a straight line; hence every straight line which passes through two of the points just mentioned, will necessarily pass through the third, and be perpendicular to the chord. PROP. E. THEOR. The arcs of a circle intercepted by two parallels are equal; and, conversely, if two straight lines intercept equal arcs of a circle, and do not cut each other within the circle, the lines will be parallel. There may be three cases: First. If the parallels are tangents to the circle, as AB, CD; then, each A Second. When, of the two parallels AB, GH, one is a tangent, the other a chord, which being perpendicular to FE, the arc GEH is bisected by FE (Cor. 4. Prop. D. Book 3.); so that in this case also, the intercepted arcs GE, EH are equal. C Third. If the two parallels are chords, as GH, JK; let the diameter FE be perpendicular to the chord GH, it will also be perpendicular to JK, since they are parallel; therefore, this diameter must bisect each of the arcs which they subtend: that is, GE=EH, and JE EK; therefore, JE-GE=EK-EH; or, which amounts to the same thing, JG is equal to HK. = Conversely. If the two lines be AB, CD, which touch the circumference, and if, at the same time, the intercepted arcs EJF, EKF are equal, EF must be a diameter (Prop. A. Book 3.); and therefore AB, CD (Cor. 3. 16. 3.), are parallel. But if only one of the lines, as AB, touch, while the other, GH, cuts the circumference, making the arcs EG, EH equal; then the diameter FE which bisects the arc GEH, is perpendicular (Schol. D. 3.) to its chord GH it is also perpendicular to the tangent AB; therefore AB, GH are parallel. If both lines cut the circle, as GH, JK, and intercept equal arcs GJ, HK; let the diameter FE bisect one of the chords, as GH: it will also bisect the arc GEH, so that EG is equal to EH; and since GJ is (by hyp.) equal to HK, the whole arc EJ is equal to the whole arc EK; therefore the chord JK is bisected by the diameter FE: hence, as both chords are bisected by the diameter FE, they are perpendicular to it; that is, they are parallel (Cor. 28 1.). SCHOLIUM. The restriction in the enunciation of the converse proposition, namely, that the lines do not cut each other within the circle, is necessary; for lines drawn through the points G, K, and J, H, will intercept equal arcs GJ, HK, and yet not be parallel, since they will intersect each other within the circle. PROP. F. PROB. To draw a tangent to any point in a circular arc, without finding the centre From B the given point, take two equal distances BC, CD on the arc; join BD, and draw the chords BC, CD: make (23. 1.) the angle CBG=CBD, and the straight line BG will be the tangent required. For the angle CBD CDB; and therefore the angle GBC (32. 3.) is also equa to CDB, an angle in the alternate segment; hence, BG is a tangent at B. G B ELEMENTS OF GEOMETRY. BOOK IV. DEFINITIONS. 1 A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. 2 In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. 3 A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumference of the circle. 4. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. 5. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. 6. A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. 7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. ODO |