the sphere ECGF, through the centre c, which will bisect the chord AB in the point (th. 41). Also, join CA, CB; and draw CD, GD, to any point D in the perimeter of the section ADB. Then, because co is perpendicular to the plane ADB, it is perpendicular both to GA and GD (def. 90). So that CGA, CGD are two right-angled triangles, having the perpendicular co common, and the two hypothenuses CA, CD, equal, being both radii of the sphere; therefore the third sides GA, GD, are also equal (cor. 2, th. 34). In like manner it is shown, that any other line, drawn from the centre G to the circumference of the section ADB, is equal to GA or GB; con. sequently that section is a circle. Scholium. The section through the centre, having the same centre and diameter as the sphere, is called a great circle of the sphere; the other plane sections being little circles. THEOREM-CXVII. F B EVERY sphere is two-thirds of its circumscribing cylinder. Let ABCD be a cylinder, circumscribing A the sphere EFGH; then will the sphere EFGH be two-thirds of the cylinder ABCD. 0 E P G For, let the plane ac be a section of the sphere and cylinder through the centre 1. Join AI, BI. Also, let FIH be parallel to AD or BC, and EIG and KL parallel to AB or DC, the base of the cylinder; the latter line KL meeting вI in м, and the circular section of the sphere in N. D H Then, if the whole plane HFBC be conceived to revolve about the line HF as an axis, the square FG will describe a cylinder AG, and the quadrant IFG will describe a hemisphere EFG, and the triangle IFB will describe a cone IAB. Also, in the rotation, the three lines or parts KL, KN, KM, as radii, will describe corresponding circular sections of those solids, namely, KL a section of the cylinder, KN a section of the sphere, and км a section of the cone. Now, FB being equal to F1 or 1G, and KL parallel to FB, then by similar triangles IK is equal to км (th. 82). And since, in the right-angled triangle IKN, IN2 is equal to Ix2 +KN2 (th. 34); and because KL is equal to the radius 1G or IN, and кMIK, therefore KL is equal to Kм2 + KN2, or the square of the longest radius, of the said circular VOL. I. 46 sections, is equal to the sum of the squares of the two others. And because circles are to each other as the squares of their diameters, or of their radii, therefore the circle described by KL is equal to both the circles described by км and KN; or the section of the cylinder, is equal to both the corresponding sections of the sphere and cone. And as this is always the case in every parallel position of KL, it follows, that the cylinder EB, which is composed of all the former sections, is equal to the hemisphere EFG and cone IAB, which are com posed of all the latter sections. But the cone IAB is a third part of the cylinder EB (cor. 2, th. 115); consequently the hemisphere EFG is equal to the remaining two-thirds; or the whole sphere EFGH equal to two-thirds of the whole cylinder ABCD. Q. E. D. Corol. 1. A cone, hemisphere, and cylinder of the same base and altitude, are to each other as the numbers 1, 2, 3. Corol. 2. All spheres are to each other as the cubes of their diameters; all these being like parts of their circumscribing cylinders. Corol. 3. From the foregoing demonstration it also appears, that the spherical zone or frustum EGNг, is equal to the difference between the cylinder EGLO and the cone INQ, all of the same common height IK. And that the spherical segment PFN, is equal to the difference between the cylinder ABLO and the conic frustum AQMB, all of the same common altitude FK. PROBLEMS. PROBLEM I. To bisect a line AB; that is, to divide it into two equal parts. From the two centres A and B, with any equal radii, describe arcs of circles, intersecting each other in c and D; and draw the line CD, which will bisect the given line AB in the point E. For, draw the radii AC, BC, AD, BD. 'Then, because all these four radii are equal, and the side CD common, the two triangles C A B ACD, BCD, are mutually equilateral: consequently they are also mutually equiangular (th. 5), and have the angle ACE equal to the angle BCE. Hence, the two triangles ACE, BCE, having the two sides AC, CE, equal to the two sides BC, CE, and their contained angles equal, are identical (th. 1), and therefore have the side AE equal to EB. Q. E. D. PROBLEM II. To bisect an angle BAC*. D From the centre A, with any radius, describe an arc, cutting off the equal lines AD, AE; and from the two centres D, E, with the same radius, describe arcs intersecting in F; then draw AF, which will B bisect the angle a as required. A * A very ingenious instrument for trisecting an angle, is described in the Mechanic's Magazine, No. 22, p. 344. For, join DF, EF. Then the two triangles ADF, AEF, having the two sides AD, DF, equal to the two AE, ef (being equal radii), and the side AF common, they are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angle BAF equal to the angle CAF. Scholium. sected. In the same manner is an arc of a circle bi PROBLEM III. Ar a given point c, in a line AB, to erect a perpendicular. From the given point c, with any radius, cut off any equal parts CD, CE, of the given line; and, from the two centres D and E, with any one radius, describe arcs intersecting in F; then join CF, which will be perpendicular as required. AD CEB Then the two For, draw the two equal radii DF, EF. triangles CDF, CEF, having the two sides CD, DF, equal to the two CE, EF, and CF common, are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the two adjacent angles at c equal to each other; there. fore the line cr is perpendicular to AB (def. 11). Otherwise. WHEN the given point c is near the end of the line. From any point D assumed above the line, as a centre, through the given point c describe a circle, cutting the given line at E; and through E and the centre D, draw the diameter EDF; then join cr, which will be the perpendicular required. AE D CB For the angle at c, being an angle in a semicircle, is a right angle, and therefore the line cr is a perpendicular (by def. 15). PROBLEM 1V. From a given point A, to let fall a perpendicular on a given line BC. From the given point A as a centre, with any convenient radius, describe an arc, cutting the given line at the two points D and E; and from the two centres D, E, with any B radius, describe two arcs, intersecting at F; then draw AGF, which will be perpendicular to BC as required. A Then the For, draw the equal radii AD, ae, and df, ef. two triangles ADF, AEF, having the two sides AD DF, equal to the two AE, EF, and AF common, are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angle DAG equal the angle EAG. Hence then, the two triangles ADG, AEG, having the two sides ad, ag, equal to the two AE, AG, and their included angles equal, are therefore equiangular (th. 1), and have the angles at & equal; consequently AG is perpendicular to BC (def. 11). Otherwise. WHEN the given point is nearly opposite the end of the line. From any point D, in the given line BC, as a centre, describe the arc of a circle through the given point A, cutting BC in E; and from the centre E, with the B radius EA, describe another arc, cutting the former in F; then draw AGF, which will be perpendicular to BC as required. D Then the For, draw the equal radii da, df, and ea, ef. two triangles DAE, DFE, will be mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angles at D equal. Hence, the two triangles DAG, DFG, having the two sides DA, DG, equal to the two DF, DG, and the included angles at D equal, have also the angles at G equal (th. 1); consequently those angles at G are right angles, and the line AG is perpendicular to DG. PROBLEM V. Ar a given point a, in a line AB, to make an angle equal to a given angle c. From the centres A and c, with any one radius, describe the arcs De, fg. Then, with radius DE, and centre F, describe an arc, cutting FG in G. Through G draw the line AG, and it will form the angle required. For, conceive the equal lines or radii, DE, FG, to be drawn. Then the two trian E D G A FB gles CDE, AFG, being mutually equilateral, are mutually equiangular (th. 5), and have the angle at a equal to the angle c. |