PROP. VIII. THEOR.

If any point be taken without a circle, from which straight lines are drawn to the circumference, whereof one passes through the centre, of those which fall upon the concave circumference, the greatest is that which passes through the centre, and of the rest, that which is nearer to the one passing through the centre is greater than one more remote; and of those which fall upon the convex circumference, the least is that between the point without the circle and the diameter, and of the rest, that which is nearer to the least is less than one more remote: and only two equal straight lines can be drawn from the given point to the circumference, one upon each side of the shortest line.

Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the circumference, whereof DA passes through the centre of those which fall upon the concave circumference AEFC, the greatest is DA which passes through the centre, and that which is nearer to it is greater than one more remote, viz. DE than DF, and DF than DC; and of those which fall upon the convex circumference GKLH, the least is DG, between the point D and the diameter AG, and that which is nearer to it is less than one more remote, viz. DK than DL, and DL than DH.

Take M the centre of the circle ABC (3. 1), and join ME, MF, MC, MH, ML, MK: Then because any two sides of a triangle are greater than the third side, therefore EM, MD are together greater than ED: but EM is equal to AM; therefore AM, MD, that is, AD, is

GB

greater than ED: Again, because EM is equal to FM, and MD common to the triangles EMD, FMD, the two sides EM, MD are equal to the two FM, MD: but the angle EMD is greater than the angle FMD; therefore (1. 24) the base ED is greater than the base FD: And, in like manner, it may be shewn that FD is greater than CD: Therefore DA is the greatest, and DE greater than DF, and DF than DC.

C

F
E

M

Again, because MK, KD are greater than MD, and MK is equal to MG, the remainder KD is greater than the remainder GD, that is, GD is less than KD: And because MK, KD are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD, therefore MK, KD are less than ML, LD (1. 21) : and MK is equal to ML; therefore the remainder KD is less than the remainder LD: And, in like manner, it may be shewn that LD is less than HD: Therefore DG is the least, and DK less than DL, and DL than DH.

Also, there can be drawn two equal straight lines from the point D to the circumference, one upon each side of the shortest line DG: For, at the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB: Then, because MK is equal to MB, and DM common to the triangles DMK, DMB, the two sides DM, MK are equal to the two DM, MB, and the angle DMK is equal to the angle DMBtherefore the base DK is equal to the base DB.

But, besides DB, no other straight line can be drawn from D to the circumference, equal to DK: For, if there can, let it be DN: Then, because DN is equal to DK, and DK to DB, therefore also DN is equal to DB,

that is, a line, nearer to the least line, is equal to one which is more remote-which is impossible.

If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle.

Let the point D be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC: the point D is the centre of the circle.

For, if not, let E be the centre; join DE, and produce it both ways to meet the circumference in F and G.

F

DE

C

Then FG is a diameter of the circle ABC: And, because in FG, a diameter of the circle ABC, there is taken the point D, which is not the centre, DG shall be the greatest line from D to the circumference, and DC greater than DB, and DB than DA (3.7): But they are likewise equal (Hyp.)—which is absurd: Therefore E is not the centre of the circle ABC: And, in like manner, it may be demonstrated, that any other point than D is not the centre: Therefore D is the centre of the circle ABC.

Wherefore, If a point &c. Q. E.D.

PROP. X. THEOR.

One circumference of a circle cannot cut another in more than two points.

If it be possible, let the circumference ABF cut the

circumference DBF in more than two points, viz. in points B, G, F.

B

D

Η

K

Take the centre K of the circle ABF, and join KB, KG, KF: Then, because within the circle DBF there is taken the point K, from which to the circumference DBF fall more than two equal straight lines KB, KG, KF, the point K (3.9) is the centre of the circle DBF: But K is

E

also the centre of the circle ABF; therefore the same point is the centre of two circles that cut one another— which is impossible (3.5).

Wherefore, One circumference &c. Q.E.D.

PROP. XI. THEOR.

If two circles touch each other internally, the straight line which joins their centres, being produced, shall pass through the point of contact.

B

Let the two circles ABC, ADE touch each other internally in the point A; and let F be the centre of the circle ABC, and G the centre of the circle ADE: the straight line which joins the centres F, G, being produced, shall pass through the point of contact, A.

F

E

For, if not, let it fall otherwise, if possible, as FGDB, and join AF, AG: Then, because AG, GF are greater than AF, that is, than BF, (for AF is equal to BF, both being from the same centre,) take away the common part FG; therefore the remainder AG is greater than the remainder BG: But AG is equal to DG; therefore DG is greater than BG, the less than the greater—which is absurd: Therefore the straight line which joins the

centres F, G cannot pass otherwise than through the point of contact A, that is, it must pass through it. Wherefore, If two circles &c. Q. E.D.

PROP. XII. THEOR.

If two circles touch each other externally, the straight line which joins their centres shall pass through the point of

contact.

B

Let the two circles ABC, ADE touch each other externally in the point A, and let F be the centre of ABC, and G the centre of ADE: the straight line which joins the points F, G, shall pass through the point of contact A.

For, if not, let it pass otherwise, if possible, as FCDG, and join AF, AG: Then, because F is the centre of the circle ABC, FA is equal to FC; and, because G is the centre of the circle ADE, GA is equal to GD: Therefore FA, AG are together equal to FC, DG, and therefore they are together less than the whole FG: But they are also together greater than it (1. 20) -which is impossible: Therefore the straight line which joins the centres F, G, cannot pass otherwise than through the point of contact A, that is, it must pass through it. Wherefore, If two circles &c. Q.E.D.

PROP. XIII. THEOR

One circle cannot touch another in more points than one, whether it touches it on the inside or outside.

For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first on the inside, in the points B, D.