1 of this. 20. I. PROPOSITION VIII. THEOREM. If fome Point be assumed without a Circle, and from it certain Right Lines be drawn to the Circle, one of which passes thro' the Centre, but the other any bow; the greatest of the Lines which fall upon the concave Part of the Circumference of the Circle, is that passing thro' the Centre; and of the others, that which is nearest to the Line, paffing thro' the Centre, is greater than that more remote. But the least of the Lines that fall upon the convex Circumference of the Circle, is that which lies between the Point and the Diameter; and of the others, that which is nigber to the leaft, is less than that which is farther diftant; and from that Point there can be drawn only two equal Lines, which shall fall on the Circumference on each Side the leaft Line, LET ABC be a Circle, out of which take any Point D. From this Point let there be drawn certain Right Lines DA, DE, DF, DC, to the Circle, whereof D A paffes thro' the Centre. I fay, DA, which paties thro' the Centre, is the greatest of the Lines falling upon A EFC, the concave Circumference of the Circle: Likewife D E is greater than D.F, and D F greater than DC. But of the Lines that fall upon HLKG the convex Circumference of the Circle, the leaft is DG, viz. the Line drawn from D, to the Diameter G A; and that which is nearest the leaft DG, is always lefs than that more remote; that is, DK is less than DL, and D L less than CH. For, find * M the Centre of the Circle ABC, and let M E, MF, MC, MH, ML, MK, be joined. Now, because AM is equal to EM; if MD, wh his common, be added. AD will be equal to EM and MD. But EM and MD are + greater than Ꭼ Ꭰ ; ED; therefore A D is alfo greater than ED. Again, because M E is equal to MF, and MD is common, then ME and MD fhall be equal to M F and MD; but the Angle EMD is greater than the Angle FMD. Therefore the Bafe ED will be + greater † 24. 1. than the Bafe F D. We prove, in the fame manner, that FD is greater than CD. Wherefore, DA is the greatest of the Right Lines falling from the Point D; DE is greater than DF, and D F is greater than DC. * Moreover, because MK and K D are greater 20.1. than M D, and M G is equal to MK; then the Remainder KD will be greater than the Remainder † Ax. 4. GD. And fo GD is lefs than KD, and confequently is the leaft. And because two Right Lines MK, KD, are drawn from M and D to the Point K, within the Triangle M L D, M K, and K D, are less than M L and LD; but MK is equal to M L. † 21. §. Wherefore the Remainder DK is less than the Remainder DL. In like manner we demonstrate, that DL is less than DH. Therefore, DG is the leaft; and DK is less than DL, and DL than D H. * I fay likewife, that from the Point D only two equal Right Lines can fall upon the Circle on each Side the leaft Line. For, make the Angle D M B at the * 23. 1. Point M, with the Right Line MD, equal to the Angle K MD, and join D B. Then, becaufe MK is equal to M B, and MD is common, the two Sides KM, MD, are equal to the two Sides M B, MD, each to each; but the Angle KM D is equal to the Angle BMD Therefore the Bafe DK is + equal † 4. 1. to the Base D B. Now I fay, no other Line can be drawn from the Point D to the Circle equal to DK; for, if there can, let it be DN. Now, fince DK is equal to D N, as alfo to D B, therefore D B fhall be equal to D N, viz. the Line drawn nearest to the leaft equal to that more remote, which has been * fhewn to by this. be impoffible. Therefore, if fome Point be affumed without a Circle, and from it certain Right Lines be drawn to the Circle, one of which passes thro' the Centre, but the others any how; the greatest of the Lines, that fall upon the concave Part of the Circumference of the Circle, is that paffing thro' the Centre; and of the others, that which is nearest to the Line, paffing thro' the Centre, is greater than that more remote. But the leaf of the Lines that fall upon tha the convex Circumference of the Circle, is that which lies between the Point and the Diameter; and of the others, that which is nigher to the leaft, is less than that which is farther diftant, and from that Point there can be drawn only two equal Lines, which shall fall on the Circumference on each Side the leaft Line; which was to be demonftrated. PROPOSITION IX. If a Point be affumed in a Circle, and from it more LE ET the Point D be affumed within the Circle ABC; and from the Point D, let there fall more than two equal Right Lines to the Circumference, viz. the Right Lines DA, DB, DC. I fay, the affumed Point D is the Centre of the Circle A B C. For, if it be not, let E be the Centre, if poffible; and join DE, which produce to G and F. * Then FG is a Diameter of the Circle ABC; and fo, because the Point D, not being the Centre of the Circle, is affumed in the Diameter F G ; therefore * 7 of this. DG will be the greatest Line drawn from D to the Circumference, and DC greater than D B, and D B than DA; but they are alfo equal, which is abfurd. Therefore E is not the Centre of the Circle A B C. And in this manner we prove, that no other Point, except D, is the Centre; therefore D is the Centre of the Circle A BC; which was to be demonstrated. +10. 1. Otherwife 2 Let ABC be the Circle, within which take the Point D, from which let more than two equal Right Lines fall on the Circumference of the Circle, viz. the three equal ones D A, D B, DC: I fay, the Point D is the Centre of the Circle ABC. For, join AB, BC, which bifect † in the Points E and Z; as alfo join ED, DZ; which produce to the the Points H, K, O, L; then, because A E is equal to EB, and ED is common, the two Sides A E, ED, fhall be equal to the two Sides BE, ED. And the Bafe D A, is equal to the Base DB: Therefore the Angle A ED will be equal to the Angle BED ;* 8. 1. and fo [by Def. 10. 1.] each of the Angles AED, BED, is a Right Angle: Therefore HK, bifecting A B, cuts it at Right Angles. And because a Right Line in a Circle, bifecting another Right Line, cuts it at Right Angles, and the Centre of the Circle is in the cutting Line, [by Cor. 1. 3.] therefore the Centre of the Circle A B C will be in HK. For the fame Reafon, the Centre of the Circle will be in OL. And the Right Lines HK, OL, have no other Point common but D: Therefore D is the Centre of the Circle A BC; which was to be demonftrated. PROPOSITION X.. THEOREM. A Circle cannot cut another Circle in more than two Points.. FOR, if it can, let the Circle ABC cut the Circle Now, because the Point K is affumed within the which is abfurd. Wherefore, a Circle cannot cut a* 5 of this, Circle in more than two Points; which was to be demonftrated. PRO. zo. 1. PROPOSITION XI. THEOREM. If two Circles touch each other on the Infide, and the Centres be found, the Line joining their Centres will fall on the [Point of] Contact of thofe Circles. LE ET two Circles A BC, A DE, touch one another inwardly in A; and let F be the Centre of the Circle A B C, and G that of ADE. I fay, a Right Line joining the Centres G and F, being produced, will fall in the Point A. If this be denied, let the Right Line, joining FG, cut the Circles in D and H. Now, because AG and F G are greater than AF, that is, than FH; take away FG, which is common, and the Remainder A G is greater than the Remainder G H. But A G is equal to GD; therefore GD is greater than GH, the lefs than the greater; which is abfurd. Wherefore, a Line drawn thro' the Points F and G, will not fall out of the Point of Contact A, and fo neceffarily must fall on it; which was to be demonftrated. PROPOSITION XII. THEOREM. If two Circles touch one another on the Outfide, a Right Line joining their Centres will pass thro the [Point of] Contact. LET two Circles ABC, ADE, touch one ano ther outwardly in the Point A; and let F be the Centre of the Circle A B C, and G that of ADE. Į fay, a Right Line drawn thro' the Centres F and G, will pass thro' the Point of Contact A. For, if it does not, let, if poffible, FCD G, fall without it, and join FA, A G. Now, fince F is the Centre of the Circle ABC, AF will be equal to FC. And becaufe G is the Centre |