† 23. 1.

† 4. I.

greatest of the Right Lines, and FD the leaft: Alfo BF is greater than F C, and FC greater than FG.

I fay, 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 becaufe 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 Bafe F G fhall bet equal to the Bafe FH. I fay, no other Right Line falling from the Point F, on the Circle, can be equal to F G. For if there can, let this be F K. Now fince FK is equal to F G, as alfo FH, FK will be equal to FH, viz. a Line drawn nigher to that paffing 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 fhall be that wherein the Center of the Circle is; the leaft, the Remainder of the fame 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 leaft or greateft Lines ; which was to be demonftrated.

PRO

PROPOSITION VIII.

THEOREM.

If fome Point be affumed without a Circle, and from it certain Right Lines be drawn to the Circle, one of which paffes thro' the Center, but the other any how; the greatest of thefe Lines, is that paffing 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 paffing 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 leaft, is less than that which is further 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.

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 paffes thro' the Center. I fay DA, which paffes through the Center, is the greateft of the Lines falling upon AEFC, the Concave Circumference of the Circle, and the leaft is DG, viz. the Line drawn from D to the Diameter GA: Likewife 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 lefs than DL, and DL lefs than D H.

For find * M the Center of the Circle A B C, and * 1 of this, 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 alfo greater than ED.

F 3

Again,

† 24. I.

20. I.

† Ax, 4.

21. 1.

23. I.

Again, because ME is equal to MF, and MD is common, then EM, MD, fhall 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 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 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 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 MLD, MK, and KD, are less than ML and LD; but MK is equal to ML. Wherefore the Remainder D K is lefs than the Remainder DL. In like Manner we demonftrate that DL is lefs than DH. Therefore DG is the leaft. And DK is less than DL, and DL than DH.

*

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 MB 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 ist equal to the Bafe DB. I fay no other Line can be drawn from the Point D to the Circle equal to DK; for if there can, let DN. Now fince DK is equal to DN, as alfo to DB, therefore DB fhall be equal to DN, viz. the Line drawn nearest to the leaft equal to that more remote, which has been fhewn to be impoffible. Therefore, if fome Point be affumed zvithout a Circle, and from it certain Right Lines be drawn to the Circle, one of which paffes through the Center, but the others any how; the greatest of thefe Lines, is that paffing 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 paffing through the Center, is greater than that

more

more remote. But the leaft 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 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 demonstrated.

PROPOSITION IX.

THEOREM.

If a Point be affumed in a Circle, and from it more than two equal Right Lines be drawn to the Circumference; then that Point is the Center of the Circle.

L

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 Center of the Circle ABC.

For if it be not, let E be the Center, if poffible, and join DE, which produce to G and F.

Then F G is a Diameter of the Circle A BC; and so because the Point D, not being the Center of the Circle, is affumed in the Diameter FG, DG will * be the greatest Line drawn from D to the Circumference, 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 demonftrated.

Otherwife

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 DA, DB, DC: I fay, the Point D is the Center of the Circle ABC.

10. 1. Euc.

#8. 1.

For join AB, BC, which bifect in the Points E and Ž; as also join ED, DZ; which produce to the Points H, K, Ó, L; then because A E is equal to EB, and ED is common, the two Sides AE, ED, fhall be equal to the two Sides BE, ED. And the Bafe DA is equal to the Bafe DB: Therefore the Angle AED will be equal to the Angle BED: And fo [by Def. 10. 1.] each of the Angles AED, BED, is a Right Angle: Therefore HK bifecting 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 fon, 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 demonftrated.

PROPOSITION X.

THEOREM.

A Circle cannot cut another Circle in more than two Points.

FO

'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 affumed 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 fhall bet the Center of the Circle DEF. By Hyp. But K is the Center of the Circle ABC. Therefore K will be the Center of two Circles cutting each other, which is abfurd. Wherefore a Circle cannot cut a Circle in more than two Points; which was to be demonftrated.

PRO