fall upon the circumference; therefore it falls within it. Wherefore, if any two points, &c. Q. E. D.

PROP. III. THEOR.

If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and, if it cut it at right angles, it shall bisect it.

Let ABC be a circle; and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: it shall cut AB at right angles.

Take* E the centre of the circle, and join EA, EB; * 1. 3. then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, the two sides AF, FE are equal to the two BF, FE, and the base EA is equal

to the base EB; therefore the angle

C

1)

* 10 Def. 1.

therefore each of the angles AFE,

BFE is a right angle; wherefore the straight line CD, drawn through the centre bisecting another, AB, that does not pass through the centre, cuts the same at right angles.

But let CD cut AB at right angles; CD shall also bisect it, that is, AF shall be equal to FB.

The same construction being made, because EA, EB from the centre are equal to one another, the angle EAF is equal to the angle EBF; and the right angle * 5. 1. AFE is equal to the right angle BFE: therefore, the two triangles EAF, EBF, have two angles in the one

equal to two angles in the other, and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are *26. 1. *equal; and therefore AF is equal to FB. Wherefore, if a straight line, &c. Q. E. D.

* 1.3.

PROP. IV. THEOR.

If in a circle two straight lines cut one another, which do not both pass through the centre, they do not bisect each other.

Let ABCD be a circle, and AC, BD two straight lines in it which cut one another in the point E, and do not both pass through the centre: AC, BD shall not bisect one another.

B

D

d

For, if it is possible, let AE be equal to EC, and BE to ED; then, if one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre: but if neither of them pass through the centre, take * F the centre of the circle, and join EF: and because FE, a straight line through the centre, bisects another AC which does not pass through the centre, it cuts it at right angles; wherefore FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, it cuts it at right* angles; wherefore FEB is a right angle: but FEA was proved to be a right angle; therefore FEA is equal to the angle FEB, the less to the greater, which is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D.

PROP. V. THEOR.

If two circles cut one another, they shall not have the

same centre.

Let the two circles ABC, CDG cut one another in the points B, C; they shall not have the same centre. For, if it be possible, let E be their centre: join EC, and draw any straight line EFG, meeting them in F and G: and because E is the centre

F

G

of the circle ABC, CE is equal to
EF. Again, because E is the centre D

of the circle CDG, CE is equal to

E

EG: but CE was proved to be equal

to EF; therefore EF is equal to EG,

B

the less to the greater, which is impossible: therefore E is not the centre of the circles ABC, CDG. Where

fore, if two circles, &c. Q. E. D.

PROP. VI. THEOR.

If one circle touch another internally, they shall not have the same centre.

Let the circle CDE touch the circle ABC internally in the point C: they shall not have the same

centre.

For, if they have, let it be F; join FC, and draw

equal to FB; therefore FE is equal

to FB, the less to the greater, which is impossible;

H

therefore F is not the centre of the circles ABC, CDE. Wherefore, if one circle, &c. Q. E. D.

PROP. VII. THEOR.

If any point be taken in the diameter of a circle which is not the centre, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line which passes through the centre is always greater than one more remote: and from the same point there can be drawn two, and only two, straight lines to the circumference, that are equal to one another, one upon each side of the diameter.

Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre: let E be the centre; of all the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA, that in which the centre is, shall be the greatest, and FD, the other part of the diameter AD, shall be the least: and of the others, FB, the nearer to FA, shall be greater than FC the more remote, and FC than FG.

B

A

Join BE, CE, GE; and because two sides of a tri* 20. 1. angle are greater than the third, therefore FE, EB are greater than FB; but EA is equal to EB; therefore FE, EA, that is, FA, is greater than FB. Again, because EB is equal to EC, and EF common to the triangles FBE, FCE, the two sides FE, EB are equal to the two FE, EC, each to each; but the angle FEB is greater than the angle FEC; therefore the base FB is

G

D

E

* 24. 1. greater than the base FC. In like manner it may be proved that FC is greater than FG. Again, be

cause EF, FG are greater than EG, and EG is equal 20. 1. to ED; therefore EF, FG are greater than ED: take away the common part EF, and the remainder FG is greater than the remainder FD. Therefore, FA is the greatest, and FD the least of all the straight lines from F to the circumference; and FB is greater than FC, and FC than FG.

Also, there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the diameter. At the point E, in the straight line EF, make the angle FEH equal to the * 23. 1. angle FEG, and join FH: then, because EG is equal to EH, and EF common to the two triangles FEG, FEH; the two sides FE, EG are equal to the two FE, EH; and the angle FEG is equal to the angle FEH; therefore the base FG is equal to the base * 4. 1. FH: but, besides FH, no other straight line can be drawn from F to the circumference equal to FG; for, if there can, let it be FK; and because FK is equal to FG, and FG to FH, FK is equal to FH; that is, a line nearer to that which passes through the centre, is equal to one which is more remote; which has been proved to be impossible. Therefore, if any point be taken, &c. Q. E. D.

PROP. VIII. THEOR.

If any point be taken without a circle, and straight lines be drawn from it 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 through the centre is always greater than one more remote: but 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 always less than one more remote: and only two equal