angle DAE (16. 1), and therefore than its equal DBE, and so the side DB is greater than DE (19. 1); but DF is equal to DB [Def. 10. 1], therefore DF is greater than DE, the part than the whole, which is absurd: therefore the right line drawn from A to B does not in any part fall without the circle. In like manner it may be proved, that no part of it falls on the circumference, it falls therefore wholly within the circle.

PROP. III. THEOR.

If a right (CD), passing through the centre of a circle (ABC), bisect a right line inscribed in it (AB), not passing through its centre, it cuts it at right angles; and if it cut it at right angles, it bisects it.

Find the centre of the circle E [1. 3], and join EA, EB, which, being radiuses of the circle, are equal [Def. 10. 1], and the triangle EAB is isosceles [Def. 29. 1]; therefore if EF or CD bisect the base AB, it cuts it at right angles, and, if it cut it at right angles, it bisects it [Cor. 26. 1].

A

PROP. IV. THEOR.

Two right lines inscribed in a circle, cutting each other, and not passing both through the centre, do not bisect each other.

If one of the right lines pass through the centre, it is manifest, it is not bisected by the other, not passing through the

centre.

B

T

E

But if neither of them, as AC, BD, pass through the centre, they cannot bisect each other; for let them, if possible, do it, and find the centre of the circle F (1. 3), and join EF; and since AC is bisected in E [Hyp.], FE is perpendicular to AC [3. 3], and the angle FEC right; and since BD is bisected in E, FE is perpendicular to BD [3. 3], and the angle FED right,

and therefore equal to FEC, the part to the whole, which is absurd. Therefore AC and BD do not bisect each other.

PROP. V. THEOR.

If two circles (ACF, DCG) cut each other, they have not the

same centre.

For, if possible, let E be the centre of both circles, and draw EC to the intersection, and EFG meeting the circles in F, G.

Because E is the centre of the circle ACF [Hyp.], EF is equal to EC [Def. 10. 1]; and. because E is the centre of the circle DCG, EG is equal to EC [Def. 10. 1]; whence EF, EG, being each equal to EC,

A

E

B

are equal to each other [Ax. 1. 1], part and whole, which is absurd; therefore E is not the centre of both the circles ACF, DCG. In like manner it may be shewn, that no other point can be their centre.

PROP. VI. THEOR.

If two circles (ACB, DCE) touch each other internally, they have not the same centre.

For, if possible, let F be the centre - of both circles, and draw FC the contact, and FEB meeting them in E and B.

Because F is the centre of the circle DCE [Hyp.], FE is equal to FC [Def. 10. 11; and because F is the centre of the circle ACB, FB is equal to FC [Def. 10. 1]; whence, FE, FB, being each equal to FC,

D

F

are equal to each other [Ax. 1. 1], part and whole, which is absurd: therefore F is not the centre of both the circles ACB, DCE. In like manner it may be shewn, that no other point can be their centre.

PROP. VII. THEOR.

If any point (D), be taken within a circle (GEA), different from the centre (C); the greatest right line which can be drawn from it to the circumference, is that (DG), which passes through the centre (C).

The remaining part (DA) of the diameter (GA), passing through the point so taken, is the least.

Of others DB, DE) drawn from that point to the circumference, the right line (DB), which is nearer to that passing through the centre, is greater than one (DE) which is more

remote.

And from that point, there can be drawn to the circumference, but two right lines (as DF, DH) equal to each other. Part 1.-DG passing through the centre, is greater than any other, as DB.

Draw CB, and CG is equal to CB [Def. 10. 11, add to each CD, and DG is equal to DC, CB together; but DC, CB together are greater than DB [20. 1], therefore DG is also greater than DB.

Part 2.-The remaining part DA of the diameter GA is less than any other, as DF.

E

B

G

H

Draw CF, and CD, DF together are greater than CF [20. 1], and therefore than its equal CA; taking from each CD which is common, DF is greater than DA [Ax. 5].

Part 3.-DB which is nearer to that DG which passes through the centre, is greater than any DE, which is more

remote.

Draw CE, and in the triangles BCD, ECD, the sides BC, CD are severally equal to the sides EC, CD, but the angle BCD is greater than the angle ECD, the whole than its part, therefore the base BD is greater than the base ED [24. 1]. In like manner ED may be proved greater than FD.

Part 4.-More than two equal right lines cannot be drawn from that point to the circumference.

For however three right lines be drawn from D to the cir cumference, either one of them is part of the diameter, and therefore greater or less than either of the others, by part ist and 2nd; or two of them are on the same part of the diameter, and therefore unequal, by part 3rd.

PROP. VIII. THEOR.

If from any point (D) without a circle, right lines be drawn to the circumference of a circle (GMA); of those drawn to the concave circumference, the greatest is that (DA) which passes through the centre (C).

of the rest, that which is nearer to that through the centre, is greater than the more remote.

But of those which fall on the convex circumference, the least is that, which, being produced, would pass through the centre. Of the rest, that which is nearer to the least, is less than the more

remote.

Only two equal right lines can be drawn from that point to the circumference.

Part 1.-Of those which fall on the concave circumference, that DA which passes through the centre, is greater than any other, as DE.

Draw CE, and CE is equal to CA (Def. 10. 1), add to each CD, and DA is equal to DC, CE together; but DC, CE together are greater than DE (20. 1), therefore DA is greater than DE.

Part 2.-That DE, which is nearer to M that DA through the centre, is greater than the more remote DF.

Draw CF, and in the triangles DCE,

E

A

DCF, the sides DC, CE are severally equal to DC, CF, and the angle DCE is greater than DCF, therefore the base DE is greater than the base DF (24. 1). In like manner DF may be proved greater than DM.

Part. 3. Of those which fall on the convex circumference, that DG, which being produced would pass through the centre, is less than any other, as DK.

Draw CG, CK; and DK, KC are greater than DC (20. 1), taking from them, the equals CK, CG, the right line DK is greater than DG (Ax. 5).

Part 4.-That DK, which is nearer to the least DG, is less than the more remote DL.

1

Draw CL; and DL, LC together, are greater than DK, KC together (21. 1); taking from them the equals CL, CK, the right line DL is greater than DK (Ax. 5. 1). In like manner it may be proved, that DL is less than DH.

Part 5.-Only two equal right lines can be drawn from D to the circumference.

For however three right lines be drawn from D to the circumference, either one of them, produced if necessary, passes through the centre, and is therefore either greater or less than either of the others, by parts 1 and 3; or two of them are on the same part of the diameter, and therefore unequal, by parts 2 and 4.

PROP. IX. THEOR.

If from any point within a circle, more than two equal right lines can be drawn to the circumference, that point is the centre of the circle.

For if it were not the centre, only two equal right lines could be drawn from it to the circumference (7.3), which is contrary to the hypothesis.

PROP. X. THEOR.

One circle (BAF) cannot cut another (BDF) in more than two

If possible, let them cut each other in more than two points, as B, G, F; find the centre K of the circle BAF (1. 3), and draw KB, KG, KF, which are equal (Def. 10. 1); therefore, in the case of figure 1, when K is within the circle BDF, K is the centre of the same circle BDF (9. 3), therefore the circles BAF, BDF, cutting each other, have a common centre, which is absurd (5. 3); therefore the circles cannot cut