IN PROPOSITION XXXIII. THEOREM XXIII. N equal circles (AIBC, EK F G), angles, wether at the centres or cir cumferences (A M C, ENG or A IC, EK G), as also the sectors (AMC m, ENG n) have the fame ratio with the arches (Am C, EnG) on which they ftand, have to one another. Hypothefis. Thefis. I. The AIBC,ÉKFG are—to one another. I. VAMC: VENG — AmC : EnG. II. The at the centers AMC,ENG & the II. VAIC: VEKG AmC: EnG. at the OAIC, EKG ftand upon the III. Sect. AMCm:Sect.ENGnAmC:EnG. arches A m C, E n G. Preparation. 1. Join the chords A C, E G. 2. In the O A I B C, draw the chords CD, D B &c, each 3. Draw MD, MB &c, also N H, NF &c. BECAUSE DEMONSTRATION, ECAUSE on one fide the cords A C, C D, DB, & on the other the cords EG, GH, HF are to one another (Prep. 2). 1. The arches Am C, Co D, D B are all equal on the one fide, as the arches En G, GH, HF are on the other. P.28. B.3. 2. Confequently, the V AM C, CMD, DMB &c, & ENG, GNH, 4. Likewife, VEN F & the arch EGHF are equimult. of VENG, But becaufe the OAIB C, EK F G are equal (Hyp. 1). Moreover, AMC being double of V A IC, & VENG double 16. It follows that AMC VENGVAIC: VEKG. Which was to be demonftrated. 11. PREP. 4. In the arches A C, C D, take the points m &o, & join Since then the two fides A M, MC are to the two fides CM, MD Pof.1. B.1. to the ACMD. P. 4. B.1. Moreover, because the arch Am C is to the arch Co D (Arg. 1). 9. The complement AIBDC of the firft is to the complement CAIBD of the second. 10. Wherefore V Am C is to VC, D. 13. The sector A M C m is to the fector CM Do. Likewise, the sector D M B is equal to each of the two foregoing 14.Therefore the sectors A MC, CMD, D MB are to one another. 16.Wherefore, the fect. A MB DC, & the arch A CDB are equimult. of the fect. A MC m, & of the arch A m C, the sec. ENFHG, & the arch E GHF are equimult. of the fect. E N G n, & of the arch E n G. But because the OAIB C, EK F G are equal (Hyp. 1). = སྙ If the arch ACD B be to the arch EG HF, the fect. AMBDC is alfo to the fect. ENF HG, as is proved by the reasoning employed in this third part of the demonstration to arg. 12 inclufively. And, if the arch ACDB be the arch EGHF, the fect. A MB DC is alfo > the fect. ENFHG, & if lefs, lefs. Since then there are four magnitudes, the two arches Am C, EnG, & the two fect. AMC m, ENG n. And of the arch Am C, & fect. A M C m, the arch ACDB & fect. AMB DC are any equimult. whatever ; & of the arch En G, & sector E N G n, the arch EGHF, & the fect. E N FHG are any equimult. whatever. And it has been proved that, if arch A C D B be >, or< the arch EGHF, fect. A M B D C is alfo >, or<the fect. E NFHG. 17.It follows, that fect. A MC: fe&t. ENG Am C : En G, Which was to be demonftrated, 111. Ax *.3. B 1. P.27. B.3. Ax.z. B.3. P.24. B.3. Ax.2. B.1. D. 5. B.5. T COROLLARY I HE angle at the center, is to four right angles, as the arch upon which it ftands, is to the circumference. For (Fig. 1), BCD: LBD to a quadrant of the O. VBCD: 4 L = BD: 0. CORALLARY II. THE arches E F, B D of unequal circles, are fimilar, if they fub tend equal angles C & C, either at their centers, or at their Ó For But BCD or VECF: 4 LBD: OBm D. Confequently, P.15. B.5. (Fig. 2). EF: OEF VECF: 4 L. } (Cor.1.) EF: OEnFBD: OB m D. P.11. B.g. CORALLARY III. · Therefore, the arches E F, BD are fimilar. Two WO rays CB, CD cut off from concentric circumferences fimilar arches EF, BD (Fig. 2). IT REMARK. T is in confequence of the proportionality eftablished in Cor. 1. that an arch of a circle (B D) is called the MEASURE of its correfpondent angle (B C D); that is of the angle at the center, fubtended by this arch; the circumference of a circle being the only curve, whofe arches, increase or diminish in the ratio of the correspondent angles, about the fame point. The whole circle is conceived to be divided into 360 equal parts, which are called DEGREES; and each of thefe degrees into 60 equal parts, called MINUTES; and each minute into 60 equal parts, called SECONDS &c. in confequence of this hypothefis, & the correspondence established between the arches, & the angles, we are obliged to conceive all the angles about a point in the fame plane (that is the fum of 4), as divided into 360 equal parts, in fuch a manner, that the angle of a degree is no more than the 360th part of 4, or the 90th of a L, & confequently, of an amplitude to be subtended by the 360th part of the circumference. Hh I. A SOLID is that which hath length, breadth and thickness. II. That which bounds a Solid is a fuperficies. III. A ftraight line (A B) is perpendicular to a plane (PL) (Fig. 1), if it be perpendicular to all the lines (CD & F E), meeting it in this plane; that is, The line (AB) will be perpendicular to the plane (PL), if it be perpendicular to the lines (CD & F E) which being drawn in the plane (PL) pass through the point (B), fo that the angles (ABC, ABD, ABE & ABF) are right angles. IV. A plane (A B) (Fig. 2) is perpendicular to a plane (P L), if the lines (DE & FG) drawn in one of the planes (as in A B) perpendicularly to the common section (A N) of the planes, are also perpendicular to the other plane (P L.). The common fection of two planes is the line which is in the two planes : as the line (AN), which is not only in the plane (A B), but alfo in the plane (PL); therefore if the lines DE & FG drawn perpendicular to A N in the plane AB are alfo perpendicular to the plane P L; the plane A B will be perpendicular to the plane P L. V. The inclination of a fraight line (A B) to a plane, (Fig. 3.) is the acute angle (A'B E), contained by the ftraight line (A B), and another (B E) drawn from the point (B), in which A B meets the plane (PL), to the point (E) in which a perpendicular (A E) to the plane (PL) drawn from any point (A) of the line (A B) above the plane, meets the fame plane. |