* 1 of this.

† 10. I.

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 Centre, but the other any bow; the greatest of the Lines which fall upon the concave Part of the Circumference of the Circle, is that paffing through the Centre; and of the others, that which is nearest to the Line, passing through the Centre, is greater than that more remote. But the leaft of the Lines that fall upon the concave Circumference of the Circle, is that which lies between the Point and the Diameter; and of the others, that which is nigher to the least, is less than that which is farther diftant; and from that Point there can be drawn only two equal Lines, which fhall fall on the Circumference on each Side the leaft Line.

LET 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 Centre. I fay, D A, which paffes through the Centre, is the greatest of the Lines falling upon A E FC, the concave Circumfe rence of the Circle: Likewife D E is greater than DF, and D F greater than D C. But of the Lines that fall upon HL KG the convex Circumference of the Circle, the leaft is DG, viz. the Line drawn from D, to the Diameter G A ; and that which is nearest the leaft D'G, is always lefs than that more remote; that is DK is less than DL, and DL lefs than C H.

For, find * M the Centre of the Circle A B C, and Jet ME, MF, MC, MH, ML, M K, be joined.

Now, becaufe A M is equal to EM; if MD, which is common, be added, A D will be equal to EM and MD. But EM and M D are + greater than

ED;

ED; therefore AD is alfo greater than ED. Again,
Ꭼ Ꭰ
becaufe M E is equal to MF, and M D is common,
then M E and MD fhall be equal to M F and MD;
but the Angle EMD is greater than the Angle
FM D. Therefore the Bafe ED will be + greater than
the Bafe F D. We prove,, in the fame manner, that
FD is greater than C D. Wherefore, DA is the
greateft of the Right Lines falling from the Point D; DE
is greater than D F, and DF is greater than DC.

+ 24...

Moreover, because MK and KD are * greater * 20. 1. than MD, and M G is equal to M K; then the Remainder K D will be greater than the Remainder † Ax. 4. GD. And fo G D 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 lefs than M L and L D ; but M K is equal to M L. ‡ 21. 1. Wherefore the Remainder DK is lefs than the Remainder D L. In like manner we demonftrate, that DL is lefs than DH. Therefore, D G, is the leaft; DG, and DK is less than D L, and D L than D H.

23. I.

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 M B at the * Point M, with the Right Line MD, equal to the Angle KMD and join D B. Then, because K M is equal to M B, and MD is common, the two Sides K M, MD, are equal to the two Sides M B, MD, each to each; but the Angle KMD is equal to the Angle B MD. Therefore the Bafe D K is equal † 4. 1. to the Base D B. Now I fay, no other Line can be drawn from the Point D to the Circle equal to DK; for, if there can, let it be DN. Now, fince DK is equal to D N, as alfo to D B, therefore D B fhail be equal to D N, viz. the Line drawn nearest to the least equal to that more remote, which has been * fhewn to by this. be impoffible. Therefore, if fome Point be affumed without a Circle, and from it certain Right Lines be drawn to the Circle, one of which passes thro' the Centre, but the other's any how; the greateft of the Lines, that fall upon the concave Part of the Circumference of the Circle, is that paffing thro' the Centre; and of the others, that which is nearest to the Line, paffing thro' the Centre, is greater than that more remote. But the leaft of the Lines that fall upon

*

the

the convex Circumference of the Circle, is that which lies between the Point and the Diameter; and of the others, that which is higher to the leaft, is lefs 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 demonAtrated.

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 Centre of the Circle.

LE

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, D B, DC I fay, the affumed Point D is the Centre of the Circle A B C.

For, if it be not, let E be the Centre, if poffible; and join D E, which produce to G and F.

Then F G is a Diameter of the Circle A B C ; and fo, because the Point D not being the Centre of the Circle, is affumed in the Diameter F G; therefore 7 of this. DG will be the greateft Line drawn from D to the Circumference, and DC greater than DB, and D B than D A'; but they are alfo equal, which is abfurd. Therefore E is not the Centre of the Circle A B C. And in this manner we prove, that no other Point, except D, is the Centre; therefore D is the Centre of the Circle A B C; which was to be demonftrated.

† 10. I.

Otherwife:

Let A B C 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: Ifay, the Point D is the Centre of the Circle A B C.

For, join A B, B C, which bitect + in the Points E and Z; as alfo join E D, DZ; which produce to

the

*

the Points H, K, O, L; then, because A E is equal to E B, and ED is common, the two Sides A E, ED, fhall be equal to the two Sides BE, ED. And the Bafe D A, is equal to the Bafe DB: Therefore the Angle A ED will be equal to the Angle BED; * 8. 1. and fo [by Def. 10. 1.] each of the Angles A ED, BED, is a Right Angle: Therefore H K, bifecting A B, cuts it at Right Angles. And because a Right Line in a Circle, bifecting another Right Line, cuts it at Right Angles, and the Centre of the Circle is in the cutting Line, [by Cor. 1. 3.] therefore the Centre of the Circle A B C will be in H K. For the fame Reafon the Centre of the Circle will be in OL. And the Right Lines HK, OL, have no other Point common but D. Therefore D is the Centre of the Circle A BC; which was to be demonftrated.

PROPOSITION X.

THEOREM.

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

FOR, if it can, let the Circle A B C cut the Circle
DEF in more than two Points, viz. in B, G, F;
and let K be the Centre of the Circle A B C, and join
K B, K G, K F.

Now, because the Point K is affumed within the
Circle D E F, from which more than two equal Right
Lines K B, KG, K F, fall on the Circumference, the
Point K fhall be + the Centre of the Circle DEF. + 9 of this.
But K is the Centre of the Circle A BC. Therefore By Hyp.
K will be the Centre of two Circles cutting each other;
which is * abfurd. Wherefore, a Circle cannot cut a * 5 of this.
Circle in more than two Points; which was to be de-
monstrated.

PRO

20. I.

PROPOSITION XI.
THEOREM.

If two Circles touch each other on the Infide, and
the Centres be found, the Line joining their
Centres will fall on the Point of] Contact of
thofe Circles.

LET two Circles ABC, ADE, touch one another

inwardly in A; and let F be the Centre of the Circle A B C, and G that of ADE. I fay, a Right Line joining the Centres G and F, being produced, will fall in the Point A.

If this be denied, let the Right Line, joining F G, cut the Circles in D and H.

Now, because AG and F G are greater than AF, * that is, than FH; take away F G, which is common, and the Remainder A G is greater than the Remainder G H. But A G is equal to GD; therefore G D is greater than G H, the lets than the greater; which is abfurd. Wherefore, a Line drawn through the Points F and G, will not fall out of the Point of Contact A, and fo neceffarily must fall on it; which was to be demonftrated.

PROPOSITION XII.

THEOREM.

If two Circles touch one another on the Outfide, a Right Line joining their Centres will pass thro' the [Point of] Contact.

LET two Circles ABC, ADE, touch one another outwardly in the Point A; and let F be the Centre of the Circle A B C, and G that of A DE. I fay, a Right Line drawn through the Centres F and G, will pafs through the Point of Contact A.

For, it it does not, let, if poffible, FCDG, fall without it, and join F A, A G.

Now, fince F is the Centre of the Circle ABC AF will be equal to F C. And becaufe G is the

Centre