8. An angle in a segment is the angle contained by two straight

lines drawn from any point in the circumference of the segment to the extremities of the straight line which is the base of the segment.

9. And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle.

10. The sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.

11. Similar segments of a circle are those in which the angles are equal, or which contain equal angles.

PROPOSITIONS.

PROPOSITION I.

PROB. To find the centre of a given circle.

Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC.

For if it be not, let, if possible, G be the centre, and join GA, GD, GB: then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB,

because they are drawn from the centre G:* therefore the angle ADG is equal (8. 1.) to the angle GDB: but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (10 Def. 1.); therefore the angle GDB is a right angle:

But FDB is likewise a (Constr.) right angle; wherefore the angle FDB is equal (1 Ax.) to the

A

angle GDB, the greater to the less, which is impossible: therefore G is not the centre of the circle ABC.

In the same manner it can be shown, that no other point but F is the centre; that is, F is the centre of the circle ABC.

be found.

Which was to

COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other.

PROPOSITION II.

THEOR. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B shall fall within the circle. For if it do not, let it fall, if possible, without, as AEB: find (1. 111.) D the centre of the circle ABC: and join AD, DB, and

produce DF, any straight line meeting the circumference AB to E:

Then, because DA is equal (15 Def. 1.) to DB, the angle DAB is equal (5. 1.) to the angle DBA ; And because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16. 1.) than the angle DAE;

But DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE:

A

E

But to the greater angle the greater side is opposite (19. 1.), DB is therefore greater than DE:

But DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impossible: therefore the straight line drawn from A to B does not fall without the circle. In the same manner it may be demonstrated that it does not fall upon the circumference ; it falls therefore within it. Wherefore, if any two points, etc.

99

Q E. D.

* Whenever the expression" straight lines from the centre or" drawn from occurs, it is to be understood that they are drawn to the circumference.

the centre

PROPOSITION 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 cuts it also at right angles.

Take (1. III.) E the centre of the circle, and join EA, EB.

Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB; therefore the angle AFE is equal (8. 1.) to the angle BFE:

But when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right (10 Def. 1.) angle; 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 also bisects it, that is, AF is equal to FB.

The same construction being made, because EA, EB from the centre are equal (15 Def. 1.) to one another, the angle EAF is equal (5. 1.) to the angle EBF; and the right angle AFE is equal to the right angle BFE: therefore, in the two triangles EAF, EBF, there are two angles in 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 equal (26. 1.); AF therefore is equal to FB. Wherefore, if a straight line, etc. Q. E. D.

PROPOSITION 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 do not bisect one another.

For, if it is possible, let AE be equal to EC, and BE to ED: 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 (1. III.) 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 shall cut it at right (3. 111.) angles; wherefore FEA is a right angle:

F

D

Again, because the straight line FE bisects the straight line BD which does not pass through the centre, it shall cut it at right (3. III.) angles; wherefore FEB is a right angle:

VOL II.

E

And FEA was shown to be a right angle: therefore FEA is equal (1 Ax.) 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, etc. Q. E. D.

PROPOSITION 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 have not 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 of the circle ABC, CE is equal (15 Def. 1.) to EF:

Again, because E is the centre of the circle CDG, CE is equal (15 Def. 1.) to EG:

C

E

B

But CE was shown to be equal to EF; therefore EF is equal to EG, the less to the greater, which is impossible: therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, etc. Q. E. D.

PROPOSITION VI.

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

Let the two circles ABC, CDE touch one another in the point C: they have not the same centre.

For if they have, let it be F; join FC, and draw any straight line FEB meeting them in E and B:

And because F is the centre of the circle ABC, CF is equal (15 Def. 1.) to FB; also, because F is the centre of the circle CDE, CF is equal (15 Def. 1.) to FE:

And CF was shown to be equal to FB; therefore FE is equal (1 Ax.) to FB, the less to the greater, which is impossible: wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, etc. Q. E. D.

PROPOSITION VII.

A

C

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 only two straight lines that are equal to one another, one upon each side of the shortest line.

Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre: let the centre be E; of all the

straight lines FB, FC, FG, etc. that can be drawn from F to the circumference, FA is the greatest, and FD, the other part of the diameter AD, is the least: and of the others, FB is greater than FC, and FC than FG.

Join BE, CE, GE:

And because two sides of a triangle are greater (20. 1.) than the third, BE, EF are greater than BF:

But AE is equal (15 Def. 1.) to EB; therefore AE, EF, that is AF, is greater than BF:

Again, because BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF; but the angle BEF is greater (9 Ax.) than the angle CEF; therefore the base BF is greater (24. 1.) than the base FC:

C

D

E

F

K

H

For the same reason, CF is greater than GF: Again, because GF, FE are greater (20. 1.) than EG, and EG, is equal to ED; GF, FE are greater than ED: take away the common part FE, and the remainder GF is greater (5 Ax.) than the remainder FD:

Therefore FA is the greatest, and FD the least, of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF.

Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD: at the point E, in the straight line EF, make (23. 1.) the angle FEH equal to the angle GEF, and join FH:

Then because GE is equal to EH, and EF common to the two triangles GEF, HEF, the two sides GE, EF are equal to the two HE, EF; and the angle GEF is equal (Constr.) to the angle HEF; therefore the base FG is equal (4. 1.) to the base 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 (1. Ax.) to FH; that is, a line nearer to that which passes through the centre is equal to one which is more remote; which is impossible. Therefore, if any point be taken, etc. Q. E. D.

PROPOSITION 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 that through the centre is always greater than the 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 the more remote: and only two equal straight lines can be drawn from the point to the circumference, one upon each side of the least.

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,