than the sum of the smaller and the distance between the centres. For, CD is less (20. 1.) than CA+AD, and for the same reason, AD) AC+ 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 than 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 triangle 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. H A 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 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 equal to CDB, an angle in .he 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. 8. Polygons of five sides are called pentagons; those of six sides, hexagons; those of seven sides, heptagons; those of eight sides, octagons ; and so on. 9. A polygon, which is at once equilateral and equiangular, is called a regular polygon. Regular polygons may have any number of sides; the equilateral triangle is one of three sides; and the square is one of four sides. LEMMA. Any regular polygon may be inscribed in a circle, and circumscribed about one. Let ABCDE, &c. be a regular polygon: describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of BC: join AO and OD. If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide; for the side OP is common: the angle OPC= OPB, being right; hence the side PC will apply to its equal PB, and the point C will fall on B; besides, from the nature of the polygon, the angle PCD=PBA; hence CD will take H the direction BA, and since CD=BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. B D E G F The distance OD is therefore equal to AO; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shown that the circle which passes through the points B, C, D, will also pass through the point E; and so of all the rest: hence the circle which passes through the points A, B, C, passes through the vertices of all the angles in the polygon, which is therefore inscribed in this circle. Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords; they are therefore equally distant from the centre (Th. 14. 3.): hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon, or the polygon circumscribed about the circle. COR. 1. Hence it is evident that a circle may be inscribed in, or circumscribed about, any regular polygon, and the circles so described have a common centre. COR. 2. Hence it likewise follows, that if from a common centre, circles can be inscribed in, and circumscribed about a polygon, that polygon is regu lar. For, supposing those circles to be described, the inner one will touch all the sides of the polygon; these sides are therefore equally distant from its centre; and, consequently, being chords of the circumscribed circle, they are equal, and therefore include equal angles. Hence the polygon is at once equilateral and equiangular; that is (Def. 9. B. IV.), it is regular. |