of intersection joined; shew when the triangle thus formed will be equilateral and when isosceles. 19. Three straight lines, not in the same plane, intersect in a point, and through their point of intersection another straight line is drawn within the solid angle formed by them; prove that the angles which this straight line makes with the first three are together less than the sum, but greater than half the sum, of the angles which the first three make with each other. 20. If two solid angles bounded by any number of plane angles, and having a common vertex, be such that one lies wholly within the other, the sum of the plane angles bounding the latter will be greater than the sum of the plane angles bounding the former. 21. Given the three plane angles which contain a solid angle. Find by a plane construction, the angle between any two of the containing planes. 22. Two of the three plane angles which form a solid angle, and also the inclination of their planes being given, to find the third plane angle. 23. Three lines not in the same plane meet in a point; if a plane cut these lines at equal distances from the point of intersection, shew that the perpendicular from that point on the plane will meet it in the center of the circle inscribed in the triangle, formed by the portion of the plane intercepted by the planes passing through the lines. 24. If two straight lines be cut by four parallel planes, the two segments intercepted by the first and second planes, have the same ratio to each other as the two segments intercepted by the third and fourth planes. III. 25. If planes be drawn through the diagonal and two adjacent edges of a cube, they will be inclined to each other at an angle equal to two-thirds of a right angle. 26. A cube is cut by a plane perpendicular to a diagonal plane, and making a given angle with one of the faces of the cube. Find the angle which it makes with the other faces of the cube. 27. Shew that a cube may be cut by a plane, so that the section shall be a square greater in area than the face of the cube in the proportion of 9 to 8. 28. Shew that if a cube be raised on one of its angles so that the diagonal passing through that angle shall be perpendicular to the plane which it touches, its projection on that plane will be a regular hexagon. 29. If any point be taken within a given cube, the square described on its distance from the summit of any of the solid angles of the cube, is equal to the sum of the squares described on its several perpendicular distances from the three sides containing that angle. 30. A rectangular parallelopiped is bisected by all the planes drawn through the axis of it. 31. In an oblique parallelopiped the sum of the squares on the four diagonals, equals the sum of the squares on the twelve edges. IV. 32. Having three points given in a plane, find a point above the plane equidistant from them." 33. Bisect a triangular pyramid by a plane passing through one of its angles, and cutting one of its sides in a given direction. 34. Given the lengths and positions of two straight lines which do not meet when produced and are not parallel; form a parallelopiped of which these two lines shall be two of the edges. 35. If a pyramid with a polygon for its base be out by a plane parallel to the base, the section will be a polygon similar to the base. 36. If a straight line be at right angles to a plane, the intersection of the perpendiculars let fall from the several points of that line on another plane, is a straight line which makes right angles with the common section of the two planes. 37. ABC, the base of a pyramid whose vertex is O, is an equilateral triangle, and the angles BOC, COA, AOB are right angles; shew that three times the square on the perpendicular from O on ABC, is equal to the square on the perpendicular, from any of the other angular points of the pyramid, on the faces respectively opposite to them. V. 38. Of all the angles, which a straight line makes with any straight lines drawn in a given plane to meet it, the least is that which measures the inclination of the line to the plane. 39. If, round a line which is drawn from a point in the common section of two planes at right angles to one of them, a third plane be made to revolve, shew that the plane angle made by the three planes is then the greatest, when the revolving plane is perpendicular to each of the two fixed planes. 40. Two points are taken on a wall and joined by a line which passes round a corner of the wall. This line is the shortest when its parts make equal angles with the edge at which the parts of the wall meet. 41. Find a point in a given straight line such that the sums of its distances from two given points (not in the same plane with the given straight line) may be the least possible. 42. If there be two straight lines which are not parallel, but which do not meet, though produced ever so far both ways, shew that two parallel planes may be determined so as to pass, the one through the one line, the other through the other; and that the perpendicular distance of these planes is the shortest distance of any point that can be taken in the one line from any point taken in the other. BOOK XIL LEMMA I. If from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half; and so on: there shall at length remain a magnitude less than the least of the proposed magnitudes. (Book x. Prop. I.) Let AB and C be two unequal magnitudes, of which AB is the greater. and from the remainder more than its half, and so on; For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, and let DE its multiple be greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH greater than its half, so on, and from the remainder AH take HK greater than its half, and until there be as many divisions in AB as there are in DE: and the divisions in DE be DF, FG, GE. and that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, therefore the remainder FD is greater than the remainder AK: and FD is equal to C, therefore Cis greater than AK; that is, AK is less than C. Q.E.D. And if only the halves be taken away, the same thing may in the same way be demonstrated. PROPOSITION I. THEOREM. Similar polygons inscribed in circles, are to one another as the squares on their diameters. Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHK1: and let BM, GN be the diameters of the circles: as the polygon ABCDE is to the polygon FGHKL, so shall the square on BM be to the square on GN. And because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, and as BA to AE, so is GF to FL: therefore the two triangles BAE, GFL having one angle in one equal to one angle in the other, and the sides about the equal angles proportionals, are equiangular : and therefore the angle AEB is equal to the angle FLG: but AEB is equal to AMB, because they stand upon the same circumference: (III. 21.) and the angle FLG is, for the same reason, equal to the angle FNG: therefore also the angle AMB is equal to FNG: and the right angle BAM is equal to the right angle GFN; (III. 31.) wherefore the remaining angles in the triangles ABM, FGN are equal, and they are equiangular to one another: therefore as BM to GN, so is BA to GF; (VI. 4.) and therefore the duplicate ratio of BM to GN, is the same with and the ratio of the polygon ABCDE to the polygon FGÍKL is the duplicate of that which BA has to GF: (VI. 20.) therefore as the polygon ABCDE is to the polygon FGHKL, so is the square on BM to the square on GN. Wherefore, similar polygons, &c. Q.E.D. PROPOSITION II. THEOREM. Circles are to one another as the squares on their diameters. Let ABCD, EFGH be two circles, and BD, FII their diameters. As the square on BD to the square on FH, so shall the circle ABCD be to the circle EFGH. For, if it be not so, the square on BD must be to the square on FII, as the circle ABCD is to some space either less than the circle EFGII, greater than it. First, if possible, let it be to a space Sless than a circle EFGH; and in the circle EFGH inscribe the square EFGH. (IV. 6.) This square is greater than half of the circle EFGH; because, if through the points E, F, G, H, there be drawn tangents to the circle, the square EFGH is half of the square described about the circle: (1.47.) and the circle is less than the square described about it; therefore the square EFGH is greater than half the circle. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, GM, HM, HN, NE; therefore each of the triangles EKF, FLG, GMH, HNE, is greater than half of the segment of the circle in which it stands; because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines EF, FG, GH, HE be completed, each of the triangles EKF, FLG, GMH, HNE is the half of the parallelogram in which it is: (I. 41.) but every segment is less than the parallelogram in which it is; wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it. Again, if the remaining circumferences be divided each into two equal parts, and their extremities be joined by straight lines, by con tinuing to do this, there will at length remain segments of the circle, which together are less than the excess of the circle EFGH above the space S; because, by the preceding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes. Let then the segments EK, KF, FL, LG, GM, MH, HN, NE be those that remain, and are together less than the excess of the circle EFGH above S: therefore the rest of the circle, viz. the polygon EKFLGMHN is greater than the space S. Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN: as therefore the square on BD is to the square on FH, so is the polygon AXBOCPDR to the polygon EKFLGMHN: (XII. 1.) |