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greatest of the Right Lines, and FD the least: Allo BF is greater than FC, and FC greater than FG.
I say, moreover, that there are only two equal Right Lines that can fall from the Point Fon ABCD, the Circumference of the Circle on each side the shortest Line FD. For at the given Point E, with the Right Line EF, make the Angle FEH equal to the Angle GEF, and join FH. Now because GE is equal to EH, and EF is common, the two Sides GE and EF, are equal to the two Sides HE and EF. But the Angle GEF, is equal to the Angle HEF. Therefore the Base F G shall bet equal to the Base FH. I say, no other Right Line falling from the Point F, on the Circle, can be equal to FG. For if there can, let this be FK. Now since FK is equal to FG, as also FH, FK will be equal to FH, viz. a Line drawn nigher to that passing thro' the Center, equal to one more remote, which cannot be. If, therefore, in the Diameter of a Circle, fome Point be taken, which is not the Center of the Circle, and from that Point certain Right Lines fall on the Circumference of the Circle, the greatest of these Lines shall be that wherein the Center of the Circle is; the least, the Remainder of the same Line. And of all the other Lines, the nearest to that which was drawn through the Center, is always greater than that more remote ; and only two equal Lines fall from the abovefaid Point upon the Circumference, on each side of the least or greateft Lines; which was to be demonstrated,
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 Center, but the other any how; the greatest of these Lines, , is that passing thro' the Center, and falling upon the Concave Part of the Circumference of the Circle; and of the others, that which is nearest to the Line paling thro' the Center 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 nigher to the least, is less than that which is furiber distant; and from that point there can be drawn only two equal Lines, which fall fall on the Circumference on each side the least Line.
ET ABC be a Circle, out of which take any L Point D. From this point let there be drawn certain Right Lines DA, DE, DF, DC, to the Circle, whereof DA passes thro' the Center. I say DA, which passes through the Center, is the greateft of the Lines falling upon AEFC, the Concave Circumference of the Circle, and the least is DG, viz. the Line drawn from D to the Diameter GA: Likewise DE is greater than DF, and D F greater than DC. But of these Lines that fall upon HLGK the Convex Circumference of the Circle, that which is nearest the least DG, is always less than that more remote; that is, DK is less than DL, and DL less than DH.
For find * M the Center of the Circle ABC, and * I gf ibis, let ME, MF, MC, MH, ML, be joined.
Now because AM is equal to ME; if MD, which is common, be added, AD will be equal to EM and MD. But EM and MD are + greater + 20 than ED, therefore AD is also greater than ED.
Again, because ME is equal to MF, and MD is common, then EM, MD, shall be equal to MF, MD; and the Angle EMD is greater than the Angle FMD. Therefore the Base ED will be + greater than the Base F D. We prove, in the same Manner that F D is greater than CD. Wherefore DA is the greatest of the Right Lines falling from the Point D; DE is greater than DF, and DF is greater than DC.
Moreover, because MK and KD are * greater than MD, and MG is equal to MK; then the Remainder KD will + be greater than the Remainder GD. And so GD 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 ML and LD; but MK is equal to ML. Wherefore the Remainder DK is less than the Remainder DL. In like Manner we demonstrate that DL is less than DH. Therefore D G is the least. And DK is less than DL, and DL than DH.
I say, likewise, that from the Point D only two equal Right Lines can fall upon the Circle on each Side the least Line. For make * the Angle DMB at the Point M, with the Right Line MD, equal to the Angle KMD, and join DB. Then because MK is equal to MB, and MD is common, the two Sides KM, MD, are equal to the two Sides BM, MD, each to each; but the Angle KMD is equal to the Angle B MD. Therefore the Base DK is + equal to the Base D B. I say no other Line can be drawn from the Point D to the Circle equal to DK; for if there can, let DN. Now since D K is equal to DN, as also to DB, therefore DB shall be equal to DN, viz. the Line drawn nearest to the least equal to that more remote, which has been shewn to 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 through the Center, but the others any how; the greatest of these Lines, is that passing through the Center, and falling upon the Concave Part of the Circumference of the Circle; and of the others that which is nearest to the Line pafing through the Center, is greater than that
But the least of the Lines that fall upon
than two equal Right Lines be drawn to the Cir-
ET the Point D be assumed 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 say the assumed Point D is the Center of the Circle ABC.
For if it be not, let E be the Center, if possible, and join DE, which produce to G and F.
Then F G is a Diameter of the Circle ABC; and so because the Point D, not being the Center of the Circle, is assumed in the Diameter FG, DG will * be the greatest Line drawn from D to the Circumfe- * 7 of this rence, and DC greater than D B, and DB than DA; but they are also equal, which is abfurd. Therefore E is not the Center of the Circle ABC. And in this Manner we prove that no other Point except D is the Center ; therefore D is the Center of the Circle ABC; which was to be demonstrated.
Let ABC be the Circle, within which take the
* 8. I.
10. 1. Eue.
For join AB, BC, which bisect * in the Points E and Ž; as also join ED, DZ; which produce to the Points H, K, O, L; then because A E is equal to E B, and ED is common, the two Sides AE, ED, shall be equal to the two Sides BE, ED. And the Base DA is equal to the Base DB: Therefore the Angle A ED will be * equal to the Angle BED: And so [by Def. 10. 1.] each of the Angles AED, BED, is a Right Angle: Therefore HK bisecting AB, cuts it at Right Angles. And because a Right Line in a Circle, bifecting another Right Line, cuts it at Right Angles, and the Center of the Circle is in the cutting Line, [by Cor. 1. 3.] the Center of the Circle ABC will be in HK. For the fame Rea.. son, the Center of the Circle will be in QL. And the Right Lines HK, OL, have no other Point common but D: Therefore D is the Center of the Circle ABC; which was to be demonstrated.
OR 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 Center 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 KB, KG, KF, fall on the Circumference, + 9 of this. the Point K shall be † the Center of the Circle DEF. | By Hyp. But K is I the Center of the Circle ABC. Therefore
K will be the Center 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.