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PROPOSITION III.
THE ORE M.

If in a Circle, a Right Line drawn thro' the Cen-
tre 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 An-
gles, it fall cut it into two equal Parts.

LE

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

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For, find E the Centre of the Circle, and let 1 of this. EA, E B, be joined.

Then, because A F is equal to F B, and F E is common, the two Sides A F, F E, are equal to the two Sides B F, F E, each to each; but the Base E A is equal to the Base E B. Wherefore the Angle A FE shall be + equal to the Angle BF E. But when a + 8. 1. Right Line ftanding upon a Right Line makes the adjacent Angles equal to one another, each of the equalAngles is a Right Angle. Wherefore A FE, or Def. 10.1 BF E, is a Right Angle. And therefore the Right Line C D 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, AF will be equal to F B. For the fame Construction remaining, becaufe EA, being drawn from the Centre, is equal to E B, the Angle E AF fhall be equal to the Angle E B F. But* the Right Angle A F E is equal to the Right Angle BFE: Therefore the two Triangles E AF, EBF, have two Angles of the one equal to two Angles of the other, and the Side E F is common to both. Wherefore the other Sides of the one fhall be + equal to the+ 26, 1, other Sides of the other: And fo A F will be equal to F B. Therefore, if in a Circle, a Right Line drawn through the Centre cuts any other Right Line not drawn through the Centre into two equal Parts, it shall cut it at Right Angels, and if it cuts it at Right Angles, it shall cut it into two equal Parts; which was to be demonftrated.

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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.

L

ET ABCD be a Circle, wherein two Right Lines A C, BD, not drawn through 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 ED. Let the Centre of this. F of the Circle A B CD be + found, and join E F.

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

PROPOSITION V.
THEOREM.

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

LET the two Circles A B C, 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, becaufe E is the Centre of the Circle ABC, CE will be equal to E F. Again, becaufe E is the Centre of the Circle to CDG, CE is equal to EG. But CE has been fhewn to be equal E F. Therefore E F fhall be equal to E G, a lefs to a greater, which cannot be. Therefore the Point E is not the Centre of both the Circles A B C, CD G. Wherefore, if two Circles cut one another, they shall not have the fame Centre; which was to be demonstrated.

PROPOSITION VI.

THEOREM.

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

L

ET two Circles A B C, C D E, 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 A B C, C F is equal to F B. And becaufe F is alfo the Centre of the Circle CDE, CF fhall be equal to F E. But CF has been fhewn to be equal to F B. 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 A B C, CDE, Wherefore, if two Circles touch one another inwardly, they will not have one and the fame Centre; which was to be demonftrated.

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20. I.

† 24. I.

20. I.

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 Cir-
cumference 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.

L

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, F C, F G, fall on the Circumference: I fay, FA is the greatest of thefe Lines, and F D the leaft; and of the others F B is greater than F C, and EC greater than F G.

- For, let B E, CE, GE, be joined.

Then, because two Sides of every Triangle are a greater than a third; BE and E F-are greater than BF. But A E is equal to B E. Therefore B E and EF are equal to AF. And fo A F is greater than FB.

Again, becaufe B E is equal to CE, and FE is common, the two Sides B E and F E are equal to the two Sides CE and E F. But the Angle B E F is greater than the Angle CEF. Wherefore the Bafe BF is greater than the Bafe F Ct. For the fame Reason, C F is greater than F G. Again, becaufe GF and FE are greater than G E, and GE is equal to ED; GF and F E fhall be greater than ED; and if F E, which is common, be taken away, then the Remainder G F is greater than the Remainder F D. Wherefore, F A is the greatest of the Right Lines, and F D the leaft: Alfo BF is greater than F C, and F C greater than F G.

I fay, moreover, that there are only two equal Right Lines that can fall from the Point Fon ABCD, the Crcumference of the Circle on each side the fhorteft Line F D. For at the given Point E, with the Right Line EF, make ‡ the Angle F E H equal ‡23. 1. to the Angle G E F, and join F H. Now because GE is equal to E H, and E F is common, the two Sides GE and E F are equal to the two Sides HE and E F. But the Angle E G F is equal to the Angle HE F. Therefore the Bafe FG fhall be + equal to the Bafe + 4. 1. 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 FK. Now fince FK is equal to F G, and FH is alfo equal to F G; therefore FK will be equal to F H, viz. a Line drawn nigher to that paffing through the Centre, equal to one more remote, which cannot be. If therefore, in * by this. the Diameter of a Circle, Jame 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 through 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; which was to be demonftrated.

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