For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it. First, let it be to a space S less than the circle EFGH; and in the circle EFGH describe the square EFGH; 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 (15. 2.) of the square described about the circle; and the circle is less than the square described about it; therefore the square EFGH is greater than half of 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, MH, ΗΝ, NE; therefore each of the triangles EKF, FLG, GMH, HNE is greater than half of the segment of the circle it stands in; because if straight lines touching the circle be drawn through the points K, L, M, N, and parallelograms upon the straight lines EF, FG, GH, HE be completed; each of the triangles EKF, FLG, GMH, ΗΝΕ shall be the half (15.2.) of the parallelogram in which it is; 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: and if these circumferences before named be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there A 000 C G will at length remain segments of the circle which together shall be less than the excess of the circle EFGH above the space S: because (6. 6.), 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 of BD is to the square of FH, so (5. 6.) is the polygon AXBOCPDR to the polygon EKFLGMHN: but the square of BD is also to the square of FH, as the circle ABCD is to the space S: therefore, as the circle ABCD is to the space S, so is (15.4.) the 255466B polygon AXBOCPDR to the polygon EKFLGMHN: but the circle ABCD is greater than the polygon contained in it: wherefore the space S is greater (20. Def. 4.) than the polygon EKFLGMHN: but it is likewise less, as has been demonstrated: which is impossible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH. In the same manner, it may be demonstrated, that neither is the square of FH to the square of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square of BD to the square of 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: therefore, inversely, as the square of FH to the square of BD, so is the space T to the circle ABCD. But as the space T is to the circle ABCD, so is the circle EFGH to some space, which must be less (20. Def. 4.) than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH. Therefore, as the square of FH is to the square of BD, so is the circle EFGH to a space less than the circle ABCD, which has been demonstrated to be impossible: therefore the square of BD is not to the square of FH, as the circle ABCD is to any space greater than the circle EFGH: and it has been demonstrated, that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH: wherefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH. Circles therefore are, &c. Q. E. D. PLANE TRIGONOMETRY. LEMMA Ι. LET ABC be a rectilineal angle; if about the point B as a centre, and with any distance BA, a circle be described, meeting BA, ВС, the straight lines including the angle ABC in A, C, the angle ABC will be to four right angles, as the arc AC to the whole circumference. (20. Def. 4.). LEMMA II. Let ABC be a plane rectilineal angle as before: about Bas a centre, with any two distances BD, BA, let two circles be described meeting BA, BC in D, E, A, C; the arc AC will be to the whole circumference of which it is an arc, as the arc DE is to the whole circumference of which it is an arc. By Lemma I. the arc AC is to the whole circumference of which it is an arc, as the angle ABC is to four right angles; and by the 1 1 same Lemma I. the arc DE is to the whole circumference of which it is an arc, as the angle ABC is to four right angles; therefore the arc AC is to the whole circumference of which it is an arc, as the arc DE to the whole circumference of which it is an arc, (15. 4.). DEFINITIONS. I. LET ABC be a plane rectilineal angle; if about B as a centre, with BA any distance, a circle ACF be described, meeting BA, BC in A, C; the arc AC may be called the measure of the angle ABC, The circumference of a circle may be supposed to be divided into 360 equal parts called degrees; and each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds, &c. And as many degrees, minutes, seconds, &c. as are contained in any arc, of so many degrees, minutes, seconds, &c. is the angle, of which that arc is the measure, said to be. Cor. Whatever be the radius of the circle of which the measure of a given angle is an arc, that arc will contain the same number of degrees, minutes, seconds, &c. as is manifest from Lemma II. III. Let AB be produced till it meet the circle again in F, the angle CBF, which, together with ABC, is equal to two right angles, may be called the Supplement of the angle ABC. IV. A straight line CD drawn through C, one of the extremities of the arc AC, perpendicular upon the diameter passing through the other extremity A, may be called the Sine of the arc AC, or of the angle ABC, of which it is the measure. COR. The Sine of a quadrant, or of a right angle, is equal to the radius. V. The segment DA of the diameter passing through A, one extremity of the arc AC, between the sine CD and that extremity, may be called the Versed Sine of the arc AC, or angle ABC. VI. A straight line AE touching the circle at A, one extremity of the arc AC, and meeting the diameter BC passing through the other extremity C in E, may be called the Tangent of the arc AC, or of the angle ABC. VII. The straight line BE between the centre and the extremity of the tangent AE, may be called the Secant of the arc AC, or angle ABC. Cor. to Def. 4. 6. 7. The sine, tangent, and secant of any angle ABC, are likewise the sine, tangent, and secant of its supplement CBF. It is manifest from Def. 4. that CD is the sine of the angle CBF. Let CB be produced till it meet the circle again in G; and it is manifest that AE is the tangent, and BE the secant, of the angle ABG or EBF, from Def. 6. 7. Cor. to Def. 4. 5. 6. 7. The sine, versed sine, tangent, and secant, of any arc which is the measure of any given angle ABC, is to the sine, versed sine, tangent, and secant, of any other arc which is the measure of the same angle, as the radius of the first is to the radius of the second. Let AC, MN be measures of the angles ABC, according to def. 1. CD the sine, DA the versed sine, AE the tangent, and BE the secant of the arc AC, according to def. 4. 5. 6. 7., and NO the sine, OM the versed sine, MP the tangent, and BP the secant of the arc MN, according to the same definitions. Since CD, NO, AE, MP are parallel, CD is to NO as the radius CB to the radius NB, and AE to MP as AB to BM, and BC or BA to BD as BN or BM to BO; and (8. 4.), DA to MO as AB to MB. Hence the corollary is manifest; therefore, if the radius be supposed to be divided into any given number of equal parts, the sine, versed sine, tangent, and secant of any given angle, will each con |