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Right Angle. Therefore the Angle FDB is equal to the Angle GDB, a greater to a lefs, which is abfurd. Wherefore G is not the Center of the Circle ABC. After the fame Manner we prove, that no other Point, unless F, is the Center. Therefore F is the Center of the Circle ABC; which was to be found.

Coroll. If in a Circle, any Right Line cuts another Right Line into two equal Parts, and at Right Angles; the Center of the Circle will be in that cutting Line.

PROPOSITION II.

THEOREM.

If any two Points be affumed in the Circumference of a Circle, the Right Line joining those two Points fball fall within the Circle.

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ET ABC be a Circle; in the Circumference of which let any two Points A, B, be affumed. I fay, a Right Line drawn from the Point A to the Point B, falls within the Circle.

For let any Point E be taken in the Right Line AB, and let DA, DE, DB, be joined.

Then because DA is equal to DB, the Angle DAB will be equal to the Angle DBA; and fince the* 5. 1. Side AE of the Triangle DAE is produced, the Angle DEB will be greater than the Angle DAE,† 16. 1. but the Angle DAE is equal to the Angle DBE; therefore the Angle DEB is greater than the Angle DBE. But the greater Side fubtends the greater Angle. Wherefore DB is greater than DE. But DB only comes to the Circumference of the Circle; therefore DE does not reach fo far. And fo the Point E falls within the Circle. Therefore, if two Points are af fumed in the Circumference of a Circle, the Right Line joining thofe two Points fhall fall within the Circle; which was to be demonftrated.

Coroll. Hence if a Right Line touches a Circle, it will touch it in one Point only.

PRO

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PROPOSITION III.

THEOREM.

If in a Circle a Right Line drawn thro' the Cen ter, cuts any other Right Line not drawn thro the Center, into equal Parts, it shall cut it al Right Angles; and if it cuts it at Right Angles, it fhall cut it into two equal Parts.

ET ABC be a Circle, wherein the Right Line CD, drawn thro' the Center, bifects the Right Line AB not drawn thro' the Center. I fay, it cuts it at Right Angles.

of this. For find E the Center of the Circle, and let EA, EB, be joined.

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Then because AF is equal to F B, and FE is common, the two Sides AF, FE are equal to the two Sides BF, FE, each to each, but the Base EA is equal to the Base EB. Wherefore the Angle AFE fhall be equal to the Angle BFE. But when a Right Line standing upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles Def. 10. 1. is a Right Angle. Wherefore AFE, or BFE, is a Right Angle. And therefore the Right Line CD drawn thro' the Center, bifecting the Right Line AB not drawn thro' the Center, cuts it at Right Angles. Now if CD cuts AB at Right Angles, I fay, it will bifect it, that is AF will be equal to FB. For the fame Conftruction remaining, because EA, being drawn from the Center, is equal to EB, the Angle EAF fhall be equal to the Angle EBF. But the Right Angle AFE is equal to the Right Angle BFE; therefore the two Triangles EAF, EBF, have two Angles of the one equal to two Angles of the other, and the Side EF is common to both. Wherefore the other Sides of the one shall be † equal to the other Sides of the other: And fo AF will be equal to FB. Therefore if in a Circle a Right Line drawn thro' the Center, cuts any other Right Line not drawn thro' the Center, into two equal Parts, it fhall cut it at Right Angles; and if it cuts it at Right Angles, it shall cut it into two equal Parts ; which was to be demonstrated.

† 26. I.

PRO

PROPOSITION IV.

THEOREМ.

If in a Circle two Right Lines not being drawn thro' the Center, cut each other, they will not cut each other into two equal Parts.

L

ET ABCD be a Circle, wherein two Right Lines AC, BD, not drawn thro' the Center, cut each other in the Point E. I fay, they do not bifect each other.

For, if poffible, let them bifect each other, fo that AE be equal to EC, and BE to ED. Let the Cen*I of this. ter F of the Circle ABCD be found, and join EF. Then because the Right Line FE drawn thro' the Center, bifects the Right Line AC not drawn thro' the Center, it will† cut AC at Right Angles. And 3 of this. fo FEA is a Right Angle. Again, because the Right Line FE bifects the Right Line BD not drawn thro' the Center, it will + cut BD at Right Angles. Therefore FEB is a Right Angle. But FEA has been fhewn to be alfo a Right Angle. Wherefore the Angle FEA will be equal to the Angle FEB, a lefs to a greater; which is abfurd. Therefore AC, BD, do not mutually bifect each other. And fo if in a Circle two Right Lines, not being drawn thro' the Center, cut each other, they will not cut each other into two equal Parts; which was to be demonftrated.

PROPOSITION V.

THEOREM.

If two Circles cut one another, they shall not have the fame Center.

LE

ET the two Circles ABC, CDG, cut each other in the Points B, C. I fay, they have not the fame Center.

For if they have, let it be E, and join E C, and draw EFG at pleasure.

F

Now

Now because E is the Center of the Circle ABC, CE will be equal to EF. Again, becaufe E is the Center of the Circle CDG, CE is equal to EG. But CE has been fhewn to be equal to EF. Therefore EF fhall be equal to EG, a lefs to a greater, which cannot be. Therefore the Point E is not the Center of both the Circles ABC, CDG. Wherefore, if two Circles cut one another, they fhall not have the fame Center; which was to be demonftrated.

PROPOSITION VI.

THEOREM.

If two Circles touch one another inwardly, they will not have one and the fame Center.

ET two Circles ABC, CDE, touch one another inwardly in the Point C. I fay, they will not have one and the fame Center.

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For if they have, let it be F, and join F C, and draw F B any how.

Then because F is the Center of the Circle ABC, CF is equal to FB. And because F is only the Center of the Circle CDE, CF fhall be equal to EF. But CF has been fhewn to be equal to FB. Therefore FE is equal to F B, a lefs to a greater; which cannot be. Therefore the Point F is not the Center of both the Circles ABC, CDE. Wherefore, if two Circles touch one another inwardly, they will not have one and the fame Center; which was to be demonstrated.

PRO

PROPOSITION VII.

THEORE M.

If 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 least, the Remainder of the fame Line. And of all the other Lines, the nearest to that which was drawn thro' the Center, is always greater than that more remote, and only two equal Lines fall from the abovefaid Points upon the Circumference, on each Side of the leaft or greatest Lines.

ET ABCD be a Circle, whofe Diameter is AD,

Center of the Circle. Let the Center of the Circle be E; and from the Point F let certain Right Lines FB, FC, FG, fall on the Circumference: I fay, FA is the greatest of these Lines, and FD the leaft; and of the others FB is greater than FC, and FC greater than FG.

For let BE, CE, GE, be joined.

Then because two Sides of every Triangle are greater than the third; BE, EF, are greater than 20. 1. BF. But AE is equal to BE. Therefore B E and EF are equal to AF. And fo AF is greater than FB.

Again, because BE is equal to CE, and FE is common, the two Sides BE and FE, are equal to the two Sides CE, EF. But the Angle BEF is greater than the Angle CEF. Wherefore the Base BF is greater than the Bafe FCt. For the fame † 24. I. Reafon, CF is greater than F G.

Again, becaufe GF and FE are * greater than 20, I. GE, and GE is equal to ED; GF and FE fhall be greater than ED; and if FE, which is common, be taken away, then the Remainder GF is greater than the Remainder FD. Wherefore FA is the

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