THE ELEMENTS GEOMETRY. BOOK XII. AXIO M. AN arch of a circle is greater than its chord; but it is lefs Book XII. than two ftraight lines drawn from the same point, to touch the circle in the extremities of the arch. IF PROP. I. THEOR *. from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half; and fo on: There fhall at length remain a magnitude lefs than the least of the propofed magnitudes. Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on; there fhall at length remain a magnitude less than C. For C may be multiplied fo, as at length to become greater than AB; let it be fo multiplied, and let DE its multiple be greater *This is Prop. I. Book X. of Euclid. A D F Book XII. greater than AB, and let DE be divided into DF, FG, GE, w each equal to C. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and fo on, until there be as many divifions in AB as there are in DE: and let the divifions in AB be AK, KH, HB; and the divifions in DE be DF, FG, GE: and be. K caufe DE is greater than AB, 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, but HK is greater than the half of HA; therefore the remainder FD B CE is greater than the remainder AK: and FD is equal to C; therefore C is greater than AK; that is, AK is lefs than C. Q. E. D. b See N. H G And if only the halves be taken away, the fame thing may in the fame way be demonstrated. E PROP. II. THEOR. VERY circle is equal to the rectangle contained by its radius, and a ftraight line equal to the half of its circumference. a Let ABCD be a circle, of which the diameter is AC, and the centre E; and in AC produced, let EF be taken equal to the a 11. 1. half of the circumference, and draw the diameter BD at right angles to AC, and complete the rectangle DF; the rectangle DF is equal to the circle ABCD. I. 72. may 9. 6. 31. 1. à 36. 1. d C If not, it is either greater or less than the circle. First, let DF be greater than the circle; then fome rectangle less than DF is equal to the circle; let this be the rectangle DH: and divide BE into fuch equal parts EK, KL, LM, MB, that each of them be less than the fourth part of HF; and draw © KN, LO, MP parallel to AC, meeting the circumference in N, O, P; and complete the rectangles KC, EN, NL, KO, OM, LP, PB: and because KE is equal to KL, the rectangle EN is equal to NL: For the fame reason, KO is equal to OM, and LP to PB; therefore the rectangles NL, OM, PB are equal to the rectangles EN, KO, LP in the circle; and the rectangle KC is therefore the excefs of the rectangles about the circle above those within it: and if the fame conftruction be made in the other three fectors, it may Z f 41. 1. g Ax. 12. F may be proved in the fame way, that all the circumfcribed rect- Book XII. angles exceed all the infcribed ones by four times the rectangle KC: but because FH is greater than four times EK, the rectangle HG is greater than four times KC; and HD being equal to the circle, is greater than the infcribed rectangles; therefore DF is greater than the infcribed rectangles, together with four times KC; that is, greater than the circumfcribed rectangles. Draw e 17. 3. PY touching the circle; and join PE; then the triangles EPY, EBY together are equal to half the rectangle contained by EB and BY, YP together: But BY, YP are greater than the arch BP; therefore the figure EBYP, and much more EBXP, is greater than the rectangle contained by EB and the half of the arch BP: In the fame manner, it may be proved, that the quadrilateral EOVP is greater than the rectangle contained by EB and the half of the arch PO, and so on: Therefore all the circumfcribed rectangles together are greater than the rectangle contained by the radius and the half of the circumference; that is, greater than the rectangle DF: and it has been proved, that they are alfo lefs; which is impoffible: Therefore the rectangle DF is not greater than the circle. Y Neither is it lefs: For, if poffible, let the circle be equal to fome rectangle DZ greater than DF: Then, if the fame conftruction be made, В XP A ML K E D W T SN it may be proved, as before, that the excefs of the circumfcribed rectangles above the inscribed ones is lefs than GZ: and the circumfcribed rectangles are greater than the circle; that is, than DZ; therefore the infcribed rectangles are greater than DF: Join BP; and the triangle EBP is lefs f than the rectangle f 41. 1. contained by EB and the half of BP; and BP is less than the g Ax. 12. arch BP; and the triangle EMP is lefs than EBP; therefore much Book XII. much more is the triangle EMP less than the rectangle contained by EB and the half of the arch BP. In the fame manner, it may be proved, that the quadrilateral EPWO is lefs than the rectangle contained by EP and the half of the arch PO; and fo on: Therefore all the infcribed rectangles are less than the rectangle contained by the radius and the half of the circumference; that is, lefs than DF: and they were proved to be greater than it; which is impoffible; therefore the rectangle DF is not less than the circle: and it was before proved, not to be greater than it; therefore DF is equal to the circle. Q. E. D. COR. 1. Hence it is manifeft, that if a space be less than a circle, rectangles can be infcribed in the circle, which together are greater than that space: and that if a space be greater than a circle, rectangles can be defcribed about the circle, which are together lefs than that space. COR. 2. And confequently, if a figure be greater than any feries of rectangles in a circle, but lefs than any series of rectangles about it; that figure is equal to the circle. C PROP. III. THEOR *. IRCLES are to one another as the squares of their diameters. Let ABCD, EFGH be two circles, and BD, FH their diameters as the fquare of BD to the fquare of FH, so is the circle ABCD to the circle EFGH. c Let the fquare of BD be to the fquare of FH, as the circle ABCD to fome space Z; this space is equal to the circle EFGH. a 11. 1. Draw the diameters AC, EG at right angles to BD, FH; and b 9.6. divide b AK into any number of equal parts in M, N, O, and EL into as many in P, Q, R; and complete the rectangles KS, MT, NV; LW, PX, QY in the circles; and let the fame conftruction be made in the rest of the fectors: and because AO, 15. 5. ER are the fame parts of OC, RG; AO is to ER, as © OC to RG; therefore the fimilar rectilineal figures upon them are prod 22. 6. portionals : and the fquares of AO, ER are fimilar figures on e.Def.6. them; and the rectangles AO, OC and ER, RG are fimilar figures on OC, RG; therefore the fquare of AO is to the fquare of ER, as the rectangle AO, OC to the rectangle ER, KG: and AO is equal to ON, and ER to RQ; and the rectf 35. 3. angle AO, OC is equal to the fquare of OV, and the rectangle ER, RG to the fquare of RY; therefore the fquare of ON is to e *This is Prop. II. Book XII. of Euclid. that that of RQ, as the fquare of OV to that of RY; therefore NO BookXII. is to RQ as OV to RY; and, by alternation, NO is to OV, as QR to RY; therefore the rectangle NV is fimilar to QY; g 16.5. therefore NV is toYQ, as the fquare of NO to the fquare of ce 1.Def. 6. RQ; that is, as the fquare of BD to the fquare of FH, be- c 15. 5. caufe ON, RQ are the fame parts of BD, FH. In the fame manner, it may be proved, that the other rectangles are to one another as the square of BD to that of FH: Wherefore all the h Э rectangles in the circle ABCD are to all the rectangles in the circle EFGH as the fquare of BD to the fquare of FH: But h 12. 5. as the fquare of BD to the fquare of FH, so is the circle ABCD to the space Z; therefore, as the circle ABCD is to the space Z, fo are the rectangles in the circle ABCD to the rectangles in k 11. 5. the circle EFGH: and the circle ABCD is greater than the rect . 2. 12. angles in it; therefore the space Z is greater than the rectangles q 14. 5. in the circle EFGH. In the fame manner, it may be proved, that the space Z is lefs than any feries of rectangles about the circle EFGH: Therefore the space Z is equal to the circle 1 2. Cor. EFGH: But the fquare of BD is to the fquare of FH, as the circle ABCD to the space Z; therefore the fquare of BD is to that of FH, as the circle ABCD to the circle EFGH, Q. E. D. COR. Hence, the circumferences of circles are to one another as their diameters. m the circle n 7. 5. For a circle is equal to the rectangle contained by the radius m 2. 12. and half the circumference; therefore the rectangle contained by BK and half the circumference ABCD is to the rectangle contained by FL and half the circumference EFGH, as the circle ABCD to the circle EFGH; that is, as the fquare of BD to 0 3. 12. that of FH, or as the square of BK to that of FL; and, alternately, as the rectangle contained by BK and half the circumference ABCD is to the fquare of BK, fo is the rectangle contained by FL and half the circumference EFGH to the square of FL: and parallelograms of the fame altitude are as their bafes P; p 1. 6. therefore the half of the circumference ABCD is to BK, as half the circumference EFGH to FL; and, alternately 8, as half the circumference E e |