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 cut 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 0, is an equila. teral triangle, and the angles BOC, COA, AOB are right angles ; shew that three times the square on the perpendicular from 0 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. 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 rourd 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 XII. LEMMA I. If from the greater of two unequal magnitudes, there be taken more than its hilf, 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. 1.) Let A B and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, D B C E 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, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in AB be AK, KH, HB; And because DE is greater than AB, but BH taken from AB is greater than its half; therefore the remainder G D is greater than the remainder HA. Again, because GD is greater than HA, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK: and FD is equal to C, therefore C is 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. PROPOSITIOX L THEOREM. Bimcilor połygans inscribed in circies, are to me analise as the squares m titi dau turs. Let ABCDE. FGHEL be two circles, and in them the similar polsz018 ABCDE. EG HEL; 200 let EM, G. be the cameters of the corres: as the prizon ABCDE as to the parzon FGHAL, 50 shall the square on BM be to the bqhare on GX. Join PE, AM. GL, FY. And because the polygon ABCDE is sinnar to tbe polygon FGHKL, the angle BAE is ecca to the angle GFL, and as B.A to A E, so is GF to FL: therefore the two triangles BAE, GFL haring one angle in one equal to one angle in the other, and the sides about ile equal angles proportionals, are equiangular: and thereiore the angle A EB is equal to the angle FLG: circumference: (IIL. 21.) and the angle FLG is, for the same reason, equal to the angle FNG: therefore also the angle A MB is equal to FNG: and the right angle BAM is equal to the right angle GFV; (II1. 31.) wherefore the remaining angles in the triangles AP, FGN are equal, and they are equiangular to one another: therefore as BI to GN, so is BA to GF; (VI. 4.) and therefore the duplicate ratio of BI to GN, is the same with the duplicate ratio of B A to GP: (v. def. 10. and v. 22.) but the ratio of the square on BM to the square on GV, is the duplicate ratio of that which BM has to GV; (vi. 20.) and the ratio of the polygon ABCDE to the polygon FGHKL is the duplicate of that which В A 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 ABCI), EFGTI be two circles, and BD, FH their diameters. As the square on BD to the square on FH, so shall the circle ABCD be to the circle EFGII. 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 EFGH, or greater than it. A First, if possible, let it be to a space S less 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 tan gents 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, NĖ; 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 trianglez EKF, FLG, GMH, HNE is the half of the parallelogram in which it is : (1. 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.) but the square on BD is also to the square on FH, as the circle ABCD is to the space S; (hyp.) therefore as the circle ABCD is to the space S, so is the polygon AXBOCPDR to the polygon EKFLĠMHN: (v. 11.) but the circle ABCD is greater than the polygon contained in it; wherefore the space S is greater than the polygon EKFLGMHN: (V. 14.) but it is likewise less, as has been demonstrated; which is impossible. Therefore the square on B D is not to the square on FH, as the circle ABCD is to any space less than the circle É FGH. In the same manner, it may be demonstrated, that neither is the square on FH to the square on BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square on BD to the square on FH, as the circle ABCD is to any space greater than the circle EFGH. For, if possible, let it be so to T, a space greater than the circle EFGH, KEN H M therefore, inversely, as the square on FH to the square on BD, so is the space T to the circle ABCD; but as the space Tis to the circle ABCD, sn is the circle EFGH to some space, which must be less than the circle ABCD, (v. 14.) because the space Tis greater, by hypothesis, than the circle EFGH; therefore as the square on FH is to the square on BD, so is the circle EFGH to a space less than the circle ABCD, which has been demonstrated to be impossible; therefore the square on BD is not to the square on FH as the circle ABCD is to any space greater that the circle EFGH: and it has been demonstrated, that neither is the square on BD to the square on FH, as the circle ABCD to any space less than the circle EFGH: wherefore, as the square on BD is to the square on FH, so is the circle ABCD to the circle EFGH. Circles, therefore, are, &c. Q.E.D. |