PROPOSITION VIII. THEOREM VII. F a point (D) be taken without a circle (BGCA), & straight lines (DA, DĖ, DF, DC,) be drawn from it to the circumference, whereof one (DA) passes thro' the center (M); of those which fall upon the concave circumference, the greatest is that (DA) which passes thro' the center; & of the rest, that (DE or DF) which is nearer to that (DA) thro' the center, is always greater than (DF or DC) the more remote: but of those (DH, DK, DL, BG,) which fall upon the convex circumference, the feast is that (DH) which produced passes thro’ the center : & of the rest, that (DK or DL) which is nearer to the least (DH) is always less than (DL or DG) the more remote: & only two equal straight lines (DK, DB,) can be drawn from the point (D) unto the circumference, one upon each side of (DH) the least. Hypothesis. Thesis. DA, DE, DF, DC. DF, DC, are drawn from this DF or DC, the more remote. center M is the least of all the straigbe lines DH, DK, DL, DG. convex part in the points H, DH, is < DL or DG the more remote. fraight each side of DH the least. 1. Preparation. DEMONSTRATION. 2. DM + MA or DA will be > DE. It is demonstrated afer the same manner that: other straight line drawn from the point D to the concave part of Which was to be demonstrated I. MF (D. 15. B. 1.), & V DME > V DMF (Ax. 8. B. 1.). 4. The base DE will be also > the base DF. P. 24. B. 1. In like manner it may be shewn that: 5. The straight line DF is > DC, & fo of all the others. 6. Consequently, the straight lines DE or DF, which is nearer the line DA, which passes thro' the center, is > DF or DC more remote. Which was to be demonstrated. II. 7. Again, the sides DK + KM of the ADKM are > the third DM. P. 20. B. 1. If the equal parts MK, MH, (D. 15. B. 1.) be taken away. 8. The remainder DK will be > DH, or DH < DK. It may be proved in the same manner, that: 9. The straight line DH is < DL, & so of all the others. 10. Consequently, the straight line DH, which produced passes thro' the center M, is the least of all the straight lines drawn from the point D to the convex part of the O BGCA. Which was to be demonstrated. III. Also, DK, MK, being drawn from the extremities D & M of the side DM of the A DLM to a point K, taken within this A (Hyp. 3.). 1. It follows, that DK + MK< DL + ML. P. 21. B. L. And taking away the equal parts MK, ML, (D. 15. B. 1.). 12. The straight line DK will be <DL. In like manner it may be shewn, that : 13. The straight line DL is < DG, & fo of all the others. 14. Consequently, the straight lines DK or DL, which are nearer the line DH, which produced passes thro' the center, are < DL or DG the more remote. Which was to be demonstrated. IV. II. Preparation 1. Make V DMB = VDMK, & produce MB 'till it meets the O. P. 23. B. 1. 2. From the point D to the point B, draw the straight line DB. Pof. 1. Then, the side DM being common to the two ADKM, DBM, the side MK = the fide MB (D. 15.B.1.), & VDMK= DMB (II. Prep.1.). 15. The base DK will be = to the base DB. 4. B. 1. But because any other straight line different from DB, is either near er the line DH or more remote from it, than DB. 16. Such a straight line will be also < or > BD ( Arg. 14.). 17. Wherefore, from the point D, only two = Itraight lines DK, DB, can be drawn, one upon each side of DH. Which was to be deinonstrated. V. P. PROPOSITION IX. THEOREM VIII. If a point (D) be taken within a circle (ABC) , from which there fall more than two equal straight lines (DA, DB, DC,) to the circumference ; that point is the center of the circle. Hypothefis. Thesis, From the point D, taken within a O ABC, The point D is the center of there fall more than two equal fraight lines the O ABC. DA, DB, DC, to the O ABC. DEMONSTRATION. If not, Some other point will be the center. Becau ECAUSE the point D is not the center (Sup.), & from this point D there tall more than two equal ttraight lines DA, DB, DC, to ihe O ABC (Hyp.). 1. It follows, that from a point D, which is not the center, there can be drawn more than two equal straight lines ; which is impossible. P. 7. B. 3. 2. Consequently, the point D is the center of the O ABC. Which was to be demonstrated, PROPOSITION X. THEOREM IX. . NE circumference of a circle (ABCEG) cannot cut another (ABFCG) in more than two points (A & B). Hypothesis. Thesis. The two O ABCEG, ABFCG, cut They cut one another only in two one a nother. points A & B. DEMONSTRATION. If not, P. 1. B. 3. Pof. 1. They cut each other in more than two points, as A, B, C, &c. Preparation. draw the rays DA, DB, DC. Because ECAUSE the point D is taken within the O ABFCG, & that more than two straight lines DA, DB, DC, drawn from this point to the circumference of the O ABFCG, are equal to one another, (Prep. 1. & D. 15. B. 1.). 1. The point D is the center of this O. But this point D being also the center of the O ABCEG (Prep. 1.). 2. It would follow, that two O ABFCG, ABCEG, which cut one ano ther, have a common center D ; which is impossible. 3. Consequently, two O ABCEG, ABFCG, cannot cut one another in more than two points. Which was to be demonstrated. P. 9. B. 3. P. 5. B. 3 I PROPOSITION XI. THEOREM X. Thesis, duced, polles tbro' the point of each other internally in A. contadt A of those two O. DEMONSTRATION. If not, The straight line which joins the centers, will fall otherwise, as Preparation. Pos. 1. BECAUSE in the A CDA, the two sides CD & DA taken toge A , Therefore, if the common part CD be taken away from both sides. Ax. 5. B.L. Ax. 8.B... 4. Wherefore, the straight line CA, which joins the centers of the O AGE, ABF, which touch each other internally, being produced, will Which was to be demonstrated. 1 |