PROPOSITION III.

THEOREM.

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

L

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

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

af

Then because AF is equal to F B, and FE is common, the two Sides A F, FE, are equal to the two Sides BF, FE, each to each; but the Bafe E A is equal to the Bafe E B. Wherefore the Angle A FE fhall be + equal to the Angle BFE. But when at 8. 1. Right Line tanding upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles is a Right Angle. Wherefore A F E, or‡Def. 10. 1. BFE, is a Right Angle. And therefore the Right Line CD drawn thro' the Centre, bifecting the Right, Line A B, not drawn thro' the Centre, cuts it at Right Angles. Now, if CD cuts A B at Right Angles, I fay, it will bifect it; that is, A F will be equal to F B. For the fame Conftruction remaining, because EA, being drawn from the Centre, is equal to E B, the Angle E AF fhall be equal to the Angle EB F. But 5. 1. the Right Angle AFE is equal to the Right Angle BFE: Therefore the two Triangles E A F, E BF, 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 thef 26. 1. other Sides of the other: And fo A F will be equal to FB. Therefore, if in a Circle, a Right Line drawn thro' the Centre cuts any other Right Line, not drawn thro' the Centre, into two equal Parts, it shall cut it at Right Angles; and if it cuts it at Right Angles, it shall cụt itinto two equal Parts; which was to be demonftrated. PRO

F 2

PROPOSITION IV.

THEOREM.

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

LET ABCD be a Circle, wherein two Right

Lines AC, BD, not drawn thro' the Centre, 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 A E be equal to E C, and B E to E D. Let the Centre t of this. F of the Circle A B C D be † found, and join E F. Then, because the Right Line F E, drawn thro' the Centre, bifects the Right Line A C, not drawn thro' 3 fibis. the Centre, it will cut AC at Right Angles. And fo FEA is a Right Angle. Again, because the Right Line F E, drawn thro' the Centre, bifects the Right Line B D, not drawn thro' the Centre, it will * cut BD at Right Angles. Therefore FEB is a Right Angle. But FEA has been fhewn to be also a Right Angle. Wherefore the Angle F EA will be equal to the Angle FE B, a lefs to a greater; which is abfurd. Therefore, A C, BD, do not mutually bisect each other. And fo, if in a Circle two Right Lines, not being drawn thro' the Centre, cut each other, they will not cut each other into two equal Parts ; which was to bę demonstrated.

PROPOSITION V.

THEORE M.

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

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

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

Now,

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

PROPOSITION VI.

THEOREM.

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

LE

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

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

Then, because F is the Centre of the Circle ABC, CF is equal to F B. And because F is also the Centre of the Circle CDE, CF fhall be equal to FE. 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 Centre of both the Circles ABC, CDE. Wherefore, if twa Circles touch one another inwardly, they will not have one and the fame Centre; which was to be demonftrated,

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PROPOSITION VII.

THEOREM.

If in the Diameter of a Circle fome Point be taken, which is not the Centre 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 Centre 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 thro' the Centre, 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 greatest Line.

LE

ET ABCD be a Circle, whofe Diameter is A D, in which affume fome Point F, which is not the Centre of the Circle. Let the Centre of the Circle be E; and from the Point F, let certain Right Lines F B, FC, F G, fall on the Circumference: I Jay, 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 and EF are greater than BF. But AE is equal to B E. Therefore BE and EF are equal to A F. And fo A F is greater than FB.

Again, becaufe BE is equal to CE, and FE is common, the two Sides B E and F E are equal to the two Sides CE and EF. But the Angle BEF is greater than the Angle C E F. Wherefore the Bafe BF is greater than the Bafe FC+. For the fame Reafon, C F is greater than F G.

Again, becaufe GF and FE are greater than 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 greatest of the Right Lines, and F D the leaft: Alfo B F is greater than FC, and FC greater than F G.

I say,

I fay, moreover, that there are only two equal Right Lines, that can fall from the Point Fon A BCD, the Circumference of the Circle on each Side the shortest Line F D. For at the given Point E, with the Right Line EF, make t the Anglè FEH equal‡ 23. 1. to the Angle G EF, and join FH. Now becaufe ĠE is equal to E H, and E F is common, the two Sides GE and EF are equal to the two Sides H E and EF. But the Angle GEF is equal to the Angle HEF. Therefore the Bafe FG fhall be + equal to the Base† 4. *. 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 FG, and F H is alfo equal to FG; therefore FK will be equal to FH, viz. a Line drawn nigher to that paffing thro' the Centre, equal to one more remote, which cannot be. If, therefore, in* by this. the Diameter of a Circle, fome Point be taken, which is not the Centre of the Circle, and from that Point certain Right Lines fall on the Circumference of the Circle, the greatest of thefe Lines fhall be that wherein the Centre 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 thro' the Centre, 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 Line; which was to be demonftrated.