to the less, which is impossible: Therefore G is not the centre of the eircle 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: Which was to be found.

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,

PROP. 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.

C

D

E

B

Take any point in AB as E; find D the centre of the circle ABC join AD, DB and DE, and let DE meet the circumference in F. Then, because DA is equal 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: Now to the greater angle the greater side is opposite (19. 1.); DB is therefore greater than DE: but BD is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D.

PROP. III. THEOR.

F

If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass. through the centre, it will cut that line at right angles ; and if it cut it at right angles, it will 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. 3.) E the centre of the circle, and join EA, EB. Then because AF is equal to FB, and FE common to the two tria g SAFE, BFE, there are two sides in the one equal to two sides in the other:

but the base EA is equal to the base EB; therefore the angle AFE is equal (8. 1.) to the angle BFE. And when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right (7. Def. 1.) angle: Therefore each of the angles AFÉ, BFE is a right angle; wherefore the straight line CD, drawn through the centre bisecting AB, which does not pass through the centre, cuts AB at right angles.

A

B

F

Again, let CD cut AB at right angles; CD also bisects AB, that is, AF is equal to FB.

The same construction being made, because the radii EA, EB are equal 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; now 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, &c. Q. E. D.

PROP. IV. THEOR.

If in a circle two straight lines cut one another, which do not 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.

A

D

F

Ꮮ

B

C

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. 3.) 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 must ent it at right (3. 3.) 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 must cut it at right (3. 3.) angles; wherefore FEB is a right angle: and FEA was shown 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 cirele, &c. Q. E. D.

PROP. V. THEOR.

If two circles cut one another, they cannot have the same

centre.

Let the two circles ABC, CDG cut one another in the points B, C ; they have not the same centre.

C

G

F

E

For, if it be possible, let E be their centre: join EC, and draw any straight line EFG meeting the circles in Fand G: and because E is the centre of the circle ABC, CE is equal to EF: Again, because E is the centre of the circle CDG, CE is equal to EG: 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, &c. Q. E.D.

D

PROP. VI. THEOR.

B

If two circles touch one another internally, they cannot have the same centre.

Let the two circles ABC, CDE, touch one another internally 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 the circles in E and B; and because F is the centre of the circle ABC, CF is equat to FB; also, because F is the centre of the circle CDE, CF is equal to FE: but CF was shown to be equal to FB; therefore FE is equal 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, &c. Q. E. D.

A

PROP. VII. THEOR,

F

C

E

B

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 passing through the centre is always greater than one more remote from it: 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, &c. that can be drawn from F to the circumference, FA is the greatest, and FD, the other part of the diame ter AD, is the least: and of the others, FB is greater than FC, and FC than FG.

B

A

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 to EB; therefore AE and 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 than the angle CEF; therefore the c base BF is greater (24. 1.) than the base FC; 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: lake away the common part FE, and the remainder GF 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 BF is greater than CF, and CF than GF.

E

H'

G

D

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 Ef, 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 to angle HEF; therefore the base FG is equal (4. 1.) to the base FH: but besides FH, no straight line can be drawn from F to the circumfer ence 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 more remote, which is 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 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 unto 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, whereof DA passes through the centre. Of those which fall upon the concave part of the circumference AEFC, the greatest is AD which passes through the centre; and the line nearer to AD is always greater than the more remote, viz. DE than DF, and DF than DČ: but of those which fall upon the convex circumference HLKG, the least is DG, between the point D and the diameter AG; and the nearer to it is always less than the more remote, viz. DK than DL, and DL than DH.

D

Take (1. 3.) M the centre of the circle ABC, and join ME, MF. MC, MK, ML, MH: And because AM is equal to ME, if MD be added to each, AD is equal to EM and MD; but EM and MD are greater (20. 1.) than ED; therefore also AD is greater than ED. Again, because ME is equal to MF, and MD common to the triangles EMD, FMD, EM, MD are equal to FM, MD; but the angle EMD is greater than the angle FMD; therefore the base ED is greater (24. 1.) than the base FD. In like manner it may be shown that FD is greater than CD. Therefore DA is the greatest; and DE greater than DF, and DF than DC.

H

M

And because MK, KD are greater (20. 1.) than MD, and MK is equal to MG, the remainder KD is greater (5. Ax.) than the remainder GD, that is GD is less than KD: And because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremities of its side MD; MK, KD are less (21. 1) than ML; LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL: In like manner, it may be shown that DL is less than DH: Therefore DG is the least, and DK less than DL, and DL than DH.

F

E

A

Also there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least: at the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB; and because in the triangles KMD, BMD, the side KM is equal to the side BM, and MD common to both, and also the angle KMD equal to the angle BMD, the base DK is equal (4. 1.) to the base DB. But, besides DB, no 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 equal to DB, DB is equal to DN; that is, the line nearer to DG, the least, equal to the more remote, which has been shown to be impossible. If, therefore, any point, &c. Q. E. D.