* e £ 2. I. Then, becaufe CA is equal and parallel to DB, and to EG4, DB Book XI. is parallel to EG, and DE to GB d; but D, E, G, B, Y, S, are points taken in each of them. Join DG, YS; then DG, d 34. 1. YS, are in one plain f; and, fince DE is parallel to GB, the 2 angle EDT is equal to BGT ; but the angle DTY is equal to the a 29. 1. angle GTS ; therefore DTY, GTS, are two triangles, having 8 15. 1. the angles YDT, DTY, equal to the two angles SGT, GTS, each to each, and the fide YD equal to GS; therefore DT is equal to TG, and YT to TS. Wherefore, &c. 8; h 16. £. IF two triangular prifms of equal altitudes, the base of one of which is a parallalelogram, and the other a triangle, and, if the parallelogram be double the triangle, the prifms are equal to each other. ; Let ABCDEF, GHKLMN, be two prifms of equal altitude let the parallelogram AF be the base of the one, and the triangle GHK the base of the other; and, if AF be double GHK, the prisms are equal. b For, compleat the folids AX, GO; then, because AF is double GHK, and the parallelogram HK double the triangle GHK"; the parallelograms AF, HK, are equal; therefore the a 41. 1. folids AX, GO, are equal ; but the half of equal things are b §1. equal; therefore the prifm GHKLMN is equal to the prism ABCDEF; for each is half the folids GO, AX. Wherefore, c 28. &c. a 6—6, c 31. 3 S IMILAR polygons infcribed in circles, are to one another as the fquares of the diameters of the circles. Let ABCDE, FGHKL, be circles, in which are infcribed the fimilar polygons ABCDE, FGHKL; let BM, GN, be the diameters of the circles; then the polygon ABCDE is to FGHKL as BM fquare is to GN square. For, Join BE, AM, GL, FN; then, because the polygons. are fimilar, the angles BAE, GFL, are equal; and BA is to AE as GF to FL; wherefore the triangles ABE, FGL., are equiangular2; that is, the angle AEB equal to FLG, and the fides about them proportional; but the angle AMB is equal to b 21. 3. AEB, and FLG to FNG : therefore the angle AMB is equal to FNG, the angle BAM to GFN, and the remaining angle ABM to FGN: Therefore the triangles AMB, GFN, are equiangular, and BM is to GN as BA is to GFd; but the triangle ABM is to the triangle FGN in the duplicate ratio of AB to GF, and the polygon ABCDE is to the polygon FGHKL in the duplicate ratio of AB to GFf; but the triangle ABM is to the triangle FGN in the duplicate ratio of BM to GN; therefore the polygon ABCDE is to the polygon FGHKL in the du8 22. 5. plicate ratio of BM to GN. Wherefore, &c. d 4. 6. e 19. 6. f 20. 6. PROP PROP. I. BOOK X. LEMMA. IF F there be two unequal magnitudes, from the greater be taken a part greater than its half, and from the remainder a part taken greater than its half; this may be done till the magnitude remaining be less than any propofed magnitude. Let AB and C be two unequal magnitudes, of which AB is the greater; from AB let a part BH be taken greater than the half, and from the remainder AH a part KH greater than its half; and fo on, till the remaining magnitude, which let be AK, be less than the affigned magnitude C. Let C be multiplied till it become greater than AB, which let be DE, and divide it into the parts DF, FG, GE, each equal to C. Then, because DE is greater than AB, and the part EG taken from it lefs than the half thereof, and the part BH greater than the half of AB, there remains DG greater than AH. Again, becaufe GD is greater than HA, and GF, half of GD, is taken from it; and if from HA be taken HK greater than the half of HA, there will remain FD greater than KA; but FD is equal to C; therefore KA is lefs than C. Which was required. C PROP. II. THEOR. IRCLES are to each other as the fquares of their diame ters. Let ABCD, EFGH, be circles, whofe diameters are BD, FH; then, as the fquare of BD is to the square of FH, so is the circle ABCD to the circle EFGH. If not, the circle ABCD will be to fome figure either lefs or greater than the circle EFGH. ་ First, let it be to a figure S, lefs than the circle EFGH, in which infcribe the fquare EFGH, which will be greater than half the circle. For, if tangents are drawn to the circle, thro' the points E, F, G, H, the fquare EFGH will be half the square defcribed about the circle; but the circle is less than the fquare described about it; therefore the fquare EFGH is greater than half the circle. Let the circumferences EF, FG, GH, HE, be bisected in the points K, L, M, N, and join EK, KF, FL, LG, GM, Book XII a 41. 1. 132 ! THE ELEMENTS Book XII GM, MH, HN, NE, and if tangents are drawn from the points K, L, M, N, and parallelograms compleated upon EF, FG, GH, and HE; then each of the triangies EKF, FLG, GMH, HNE, will be equal to half the parallelogram, and therefore greater than half the segment of the circle it ftands in; if the remaining segments are bifected, and triangles drawn, as before and this be continued till the fegments are less than the excess by which the circle EFGH exceeds the figure S. Let these be the fegments cut off by the right lines EK, KF, FL, LG, GM, MH, HN, NE; then the remaining polygon EKFLGMHN will be greater than the figure S. b Lem. CI. d hyp. e II. 5. f 14. 5. ; Defcribe the polygon AXBOCPDR, in the circle ABCD, fimilar to the polygon EKFLGMHN ; then, as the fquare of BD is to the fquare of FH, fo is the circle ABCD to the figure Sd; and, as the polygon AXBOCPDR to the polygon EKFLGMHN, fo is the circle ABCD to the figure S; but the circle ABCD is greater than the polygon in it; therefore the figure S is greater than the polygon EKFLGMHNf; but it is lefs; which is abfurd; therefore the fquare of BD to the fquare of FH is not as the circle ABCD to some figure less than the circle EFGH. In like manner, it is proved that the fquare of FH to the fquare of BD is not to the circle EFGH as fome figure lefs than the circle ABCD. Laftly, the fquare of BD to the square of FH, is not as the circle ABCD to some figure greater than the circle EFGH. For, if poffible, let it be to the figure T, greater than the circle EFGH; then, inversely, the fquare of FH is to the fquare of BD as the figure T is to the circle ABCD; but, because T is greater than the circle EFGH, T will be to the circle ABCD as the circle EFGH is to fome figure less than the circle ABCD, which is proved impoffible; therefore the fquare of BD to the fquare of FH is not as the circle ABCD to fome figure less than the circle EFGH, nor to one greater; therefore, as the fquare of BD is to the fquare of FH, fo is the circle ABCD to the circle EFGH. Wherefore, &c. E PRO P. III. THE OR. VERY pyramid having a triangular base, may be divided into two pyramids, equal and fimilar to one another, having triangular bases, and fimilar to the whole pyramid; and into two equal prifms; which two prifms are greater than the half of the whole pyramid. Let |