* 1 of this.

+8. 1.

05. 1.

+ 26. 1.

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

PROPOSITION. III.

THEOREM.

If in a Circle a Right Line drawn through the Center, cuts any other Right Line not drawn through the Center, 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 Center, bifects the Right Line A B not drawn thro' the Center. I fay, it cuts it at Right Angles.

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

Then because AF is equal to FB, and FE is common, the two Sides AF, FE are equal to the two Sides BF, FE, each to each, but the Bafe E A is equal to the Bafe EB. Wherefore the Angle A FE fhall be equal to the Angle BF E. But when a Right Line ftanding upon a Right Line makes the adjacent Angles equal to one another, each of the equal Angles isa 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 Construction remaining, becaufe EA, being drawn from the Center, is equal to EB, the Angle E AF 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 shall 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.

PROPOSITION IV.

THEOREM.

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.

LE

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 Center F of the Circle ABCD be * found, and join E F. * 1 of this 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, becaufe the Right Line F E bifects the Right Line BD not drawn thro' the Center, it will † cut BD at Right Angles. Theres. fore FEB is a Right Angle. But FEA has been shewn to be also a Right Angle. Wherefore the Angle FEA will be equal to the Angle FEB, a less 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 demonstrated.

PROPOSITION. V,

THEOREM.

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

LET 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 24

Now

Now because E is the Center of the Circle CDG, CE will be equal to EF. Again, because E is the Center of the Circle CDG, CE is equal to E G. 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 shall not have the fame Center; which was to be demonstrated.

PROPOSITION VI.

THEOREM.

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

LE

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

For if they have, let it be F, and join FC, and draw FB any how.

Then because F is the Center of the Circle ABC, CF is equal to FB. And becaufe 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 FB, 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.

THEOREM.

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 thefe 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 Point upon the Circumference, on each Side of the leaft or greateft Lines.

LE

ET ABCD be a Circle, whofe Diameter is AD, in which affume fome Point F, which is not the 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 F C, and FC greater than F G.

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. I. BF. But AE is equal to B É. Therefore BE and EF are equal to AF. And fo A F is greater than FB.

Again, because B E is equal to C E, and FE is common, the two Sides BE and FE, are equal to the two Sides C E, E F. But the Angle B E F is greater than the Angle CE F. Wherefore the Bafe BF is greater than the Base F C †. Reason, CF is greater than F G.

For the fame + 24. 1.

Again, becaufe G F and F E are † greater than GE, and G E is equal to ED; GF and FE fhall be greater than ED; and if FE, which is common, be taken away, then the Remainder G Fis greater than the Remainder FD. Wherefore FA is the greatest of the Right Lines, and F E the leaft: Alfo BF is greater than FC, and FC greater than FG.

F 3

I fay,

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 F D. For at the given Point E, # 23., with the Right Line E F, make the Angle FEH equal to the Angle G É F, and join F H. Now because G E 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 GEF, is equal to the Angle HEF. Therefore the Bafe FG fhall

be

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 F K is equal to FG, as alfo FH, FK will be equal to F H, 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 thefe 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 thro' 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 greatest Lines; which was to be demonstrated.

PRO