IF PROPOSITION VIII. THEOREM VII. F a point (D) be taken without a circle (BGCA), & ftraight lines (DA, DE, DF, DC,) be drawn from it to the circumference, whereof one (DA) paffes thro' the center (M); of those which fall upon the concave circumference, the greatest is that (DA) which passes thro' the center; & of the reft, 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 thofe (DH, DK, DL, DG,) which fall upon the convex circumference, the least 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 lefs than (DL or DG) the more remote: & only two equal ftraight lines (DK, DB,) can be drawn from the point (D) unto the circumference, one upon each fide of (DH) the least. Hypothefis. 1. The point D is taken without a O BGCA in the fame plane. II. The ftraight lines DA, DE, DF, DC, are drawn from this point to the concave part of the BGCA. III. And thofe ftraight lines cut the convex part in the points H, K, L, G. THE Thefis. 1. DA which passes thro' the center M is II. DE or DF, which is nearer to DA is > III. DH which when produced paffes thro IV. DK or DL. which is nearer to the line I. Preparation. Draw the rays ME, MF, MC, MK, ML. DEMONSTRATION. HE two fides DM+ME of the A DME are > the third DE. P. 20. B. 1. And because ME = MA (D. 15. B. 1.). 2. DM + MA or DA will be > DE. It is demonstrated after the fame manner that: 3. The ftraight line DA, which paffes thro' the center M, is > any Moreover, DM being common to the two ADME, DMF, ME = In like manner it may be fhewn that: 5. The ftraight line DF is > DC, & fo of all the others. 6. Confequently, the ftraight lines DE or DF, which is nearer the line DA, which paffes thro' the center, is > DF or DC more remote. Which was to be demonftrated. II. P. 24. B. 1. 7. Again, the fides DK + KM of the A DKM 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 fame manner, that: 9. The ftraight line DH is < DL, & fo of all the others. 10. Confequently, the ftraight line DH, which produced paffes thro' the center M, is the leaft of all the ftraight lines drawn from the point D to the convex part of the O BGCA. Which was to be demonftrated. III. Alfo, DK, MK, being drawn from the extremities D & M of the fide 11. It follows, that DK + MK < DL + ML. 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 fhewn, that: 13. The ftraight line DL is < DG, & fo of all the others. 14. Confequently, the ftraight lines DK or DL, which are nearer the Which was to be demonftrated. IV. P. 21. B. L. Pof. 1. 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 ftraight line DB. Then, the fide DM being common to the two A DKM, DBM, the fide MK the fide MB (D. 15. B. 1.), & \ DMK= √ DMB (II. Prep.1.). 15. The bafe DK will be to the base DB. But because any other straight line different from DB, is either nearer the line DH or more remote from it, than DB. 16. Such a ftraight line will be alfo <or > BD (Arg. 14.). 17. Wherefore, from the point D, only two = ftraight lines DK, DB, can be drawn, one upon each fide of DH. Which was to be demonftrated. V. Q P. 4. B. 1. IF PROPOSITION IX. C THEOREM VIII Fa point (D) be taken within a circle (ABC), from which there fall more than two equal ftraight lines (DA, DB, DC,) to the circumference; that point is the center of the circle. BECAU Thefis. The point D is the center of the ABC. AUSE the point D is not the center (Sup.), & from this point D there fall more than two equal ftraight lines DA, DB, DC, to the O ABC (Hyp.). 1. It follows, that from a point D, which is not the center, there can ABC. Which was to be demonftrated, P. 7. B. 3. ON PROPOSITION X. THEOREM IX. NE circumference of a circle (ABCEG) cannot cut another (ABFCG) If not, Thefis. They cut one another only in two points A&B. DEMONSTRATION. They cut each other in more than two points, as A, B, C, &c. Preparation. 1. Find the center D of the O ABCEG. 2. From the center D to the points of fection A, B, C, &c. BECAUSE ECAUSE the point D is taken within the O ABFCG, & that more than two ftraight lines DA, DB, DC, drawn from this point to the circumference of the O ABFCG, are equal to one another, (Prep. I. & D. 15. B. 1.). 1. The point D is the center of this . But this point D being alfo the center of the ABCEG (Prep. 1.). 2. It would follow, that two O ABFCG, ABCEG, which cut one another, have a common center D; which is impoffible. P. 1. B. 3. Pof. 1. P. 9. B. 3. P. 5. B. 3. 3. Confequently, two O ABCEG, ABFCG, cannot cut one another in more than two points, Which was to be demonstrated. Ꮮ PROPOSITION XI. THEOREM X. F two circles touch each other internally in (A); the ftraight line which joins their centers being produced, shall pafs thro' the point of cortact (A). Hypothefis. The ftraight line CA joins the centers of the two AGE, ABF, which touch each other internally in A. Thefis. This fraight line CA being produced, paffes thro' the point of contact A of those two O. DEMONSTRATION. If not, The ftraight line which joins the centers, will fall otherwise, as Preparation. From the centers C & D to the point of contact A, draw the BECA USE toge USE in the A CDA, the two fides CD & DA taken to ther, are the third CA (P. 20. B. 1.), & that CA = CB (D. 15. B. i.). 1. The ftraight lines CD +DA will be alfo > CB. Therefore, if the common part CD be taken away from both fides. 2. The ftraight line DA will be > DB. But the ftraight line DA being to DG (Prep. & D. 15. B. 1.). 3. DG will be alfo > DB, which is impoffible. 4. Wherefore, the ftraight line CA, which joins the centers of the AGE, ABF, which touch each other internally, being produced, will pafs thro' the point of contact A. Which was to be demonftrated. Pof. 1. Ax. 5. B. L. Ax. 8. B. 1. |