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THEOREM. If some Poini be assumed without a Circle, and froin 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 wbich fall upon the concave Part of the Cir. cumference of the Circle, is that pasing through the Centre; and of the others, that which is nearest to the Line, passing through the Centre, is greater than that more remote. But the least of the Lines that fall upon the concave Circumference of tbe Circle, is ibai which lies between the Point and the Diameter; and of the others, that which is nigber to the least, is less than that which is fartber distant ; and from that Point there can be drawn only two equal Lines, which fall fall on the Circumference on each Side ibe least Line.
ET 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 DA passes thro' the Centre." I say, D A, which passes through the Centre, is the greatest of the Lines falling upon A EFC, the concave Circumlee sence of the Circle : Likewise DE is greater than DF, and D F greater than D C. But of the Lines that fall upon HLKG the convex Circumference of the Circle, the least is DG, viz. the Line drawn from D, to the Diameter GA; and that which is nearest the leált D'G, is always less than that more remote ; that is D K is less than DL, and D L less than CH.
For, find * At the Centre of the Circle A B C, and let ME, MF, MC, MH, ML, M K, be joined.
Now, because A M is equal to EM; if MD, which is common, be added, A D will be equal to E M and MD. But EM and MD are t greater than
ED; therefore A D is also greater than ED. Again, because M E is equal to MF, and M D is common, then M E and M D shall be equal to M F and MD; but the Angle E M D is greater than the Angle FM D. Therefore the Base E D will be + greater than
† 24. 1. the se FD. We prove,, in the same manner, that FD is greater than CD. Wherefore, D A 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 KD are *
greater than MD, and MG is equal to MK; then the Re. mainder K D will + be greater than the Remainder + Ax. 4. GD. And so G D is less than KD, and confequently is the least. And because two Right Lines MK, KD, are drawn from M and D to the Point K, within the Triangle MLD, MK, and KD, are I less than M L and LD; but M K is equal to ML. (21.3. Wherefore the Remainder D K is less than the Re. mainder D L. In like manner we demonstrate, that DL is less than DH. Therefore, DG, is the least; and D K is less than D L, and D L than D H.
I say likewise, that from the Point D only two equal Right Lines can fall
the Circle on each side the least Line. For, make * the Angle D M B at the * 23. 1. Point M, with the Right Line MD, equal to the An. gle KMD and join D B. Then, because K M is equal to M B, and MD is common, the two Sides KM, MD, are equal to the two Sides MB, MD, each to each; but the Angle KMD is equal to the Angle B M D. Therefore the Bale D K is + equal † 4. 1. to the Base D B. Now I say, 110 oiher Line can be drawn from the Point D to the Circle equal to DK; for, if there can, let it be DN. Now, lince D K is equal to D N, as also to D B, therefore D B Thail be equal to D N, viz. the Line drawn nearest to the least equal to that more remote, which has been * shewn to * by rbi:. be impossible. Therefore, if some Point be assumed without a Circle, and from it certain Right Lines be drawn to the Circle, one of which passes thro' the Centre, bui ube others any how; the greatest of the Lines, that 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, paling thro' the Centre, is greater than that more remote. Bui ihe'least of the Lines that fall upon
the convex Circumference of the Circle, is that which lies
than two equal Right Lines be drawn to the
ET the Point D be assumed within the Circle
A BC; and from the Point D lui shere fall more than two equal Right Lines to the Circumference, viz. the Right Lines DA, DB, DC liay, the allumed Point D is the Centre of the Circle A B C.
For, if it be nut, lei E be the Centre, if possible ; and join D Е, which produce to G and F.
Then F G is a Diameter of che Circle A BC; and so, because the Point D not being the Centre of
the Circle, is assumed in the Diameter FG; therefore * 7 of rbis. D G will * be the greatest Line drawn from D to the
Circumference, and D C greater than DB, and D B
For, join A B, BC, which bitect t in the Points
the Points H, K, O, L; then, because A E is equal to E B, and E D is common, the two Sides A E, ED, Ihall be equal to the two Sides B E, E D. And the Base D A, is equal to the Bale D B: Therefore the Angle A ED will be * equal to the Angle B E D; *8. 1. and fo [by Def. 10. 1.] each of the Angles A ED, BED, is a Ripht Angle: Therefore HK, bifecting A B, cuts it at Right Angles. And because a Right Line in a Circle, bisecting 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 same Reason the Centre of the Circle will be in O L. And the Right Lines HK, OL, have no other Point common but D. Therefore D is tbe Centre of the Circle A BC; which was to be demonstrated.
A Circle cannot cut another Circle in more than
FOR if it can, let the Circle ABC cut the Circle
DEF in more than two Points, viz. in B, G, F; and let K be the Centre of the Circle ABC, and join KB, KG, KF.
Now, because the Point K is assumed within the Circle DEF, from which more than two equal Right Lines K B, KG, K F, fall on the Circumference, the Point K shall be + the Centre of the Circle DEF. + 9 of tbis. But K is the Centre of the Circle A BC. Therefore I By Hyp. K will be the Centre of two Circles cutting each other; which is * absurd. Wherefore, a Circle cannot cut a * Circle in more than two Points; which was to be demonstrated.
the Centres. be found, the Line joining their
LET two Circles A B C A DE, touch one another
inwardly in A ; and ler.F be the Centre of the Circle A BC, and G that of A D E. I say, 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 A G and F G are greater than AF, * that is, than FH; take away F G, which is common, and the Remainder A G is greater than the Remainder GH. But A G is equal to GD; therefore G D is greater than GH, che lets than the greater ; which is absurd. Wherefore, a Line dı awna through the Points F and G, will not fall out of the Point of Contact A, and fo necesarily mul fall on it, which was to be demonstrated.
Right Line joining their Cenires will pass thro'
ther outwardly in the Point A; and let F be the Centre of the Circle ABC, and G that of A DE. I Pay, a Right Line drawn through the Centres F and G, will pass through the Point of Contact A.
For, it it does not, let, if possible, FCDG, fall without it, and join FA, A G.
Now, fince F is the Centre of the Circle A B C; AF will be equal to FC. And because G is the