*

D

F

LB

1. 3.

+ 15 Def. 1. 5.1.

For if it do not, let it fall, if possible, without, as AEB: find* D the centre of the circle ABC; and join DA, DB; in the circumference AB take any point F, join DF, and produce it to E: then because DA is equal to DB, the angle DAB is equal to the angle DBA: and because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the angle DAE: but DAE was proved 16. 1. equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: but to the greater angle the greater side is opposite, therefore DB is 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; 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.

*

E

19. 1.

15 Def. 1.

+ 15 Def. 1.

* 8. 1.

Take * E the centre of the circle, and join EA, EB. *1. 3. Then, because AF is equal to FB, and FE common † Hyp. to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, each to each; and the base EA is equal + to the base EB; therefore the angle AFE is equal to the angle BFE: but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of them is a right angle; therefore each of the angles AFE, BFE, 10 Def. 1. 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.

*

A

B

D

5.1.

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, +15 Def. 1. from the centre are equal + to one another, the angle EAF is equal to the angle EBF; and the right angle +10 Def. 1. AFE is equal to the right angle BFE: therefore, in the two triangles EAF, EBF, there are two angles in the one equal to two angles in the other, each to each; 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; therefore AF is equal to FB. Wherefore, if a straight line, &c. Q. E. D.

26. 1.

* 1. 3.

+ Hyp.

* 3. 3.

+ Hyp.

❤3.3.

† 1 Ax.

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.

A

BE

F

For, if it be 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 * 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 shewn 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 of the circle ABC, EC is equal to EF: again, because E is the centre of the circle CDG, EC is equal+ to EG: but EC was shewn to be equal to EF; therefore EF is equal to EG, the less to the greater, which is impossible. Therefore

A

D

E

G

E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D.

PROP. VI. THEOR.

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

Let the two circles ABC, CDE touch one another internally in the point C: they shall not have 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, FC is equal + to FB; also, because F is the centre of the circle CDE, FC is equal to FE: but FC was shewn to be equal to FB; therefore FE is equal + to FB, the less to the

A

† 15 Def. 1.

† 15 Def. 1.

† 1 Ax.

† 15 Def. 1.

F

B

† 15 Def. 1.

greater, which is impossible: therefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &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 only two straight lines that are equal to one another, one upon each side of the shortest line.

† 1 Ax.

* 20. 1.

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 shall be the greatest, and FD, the other part of the diameter AD, shall be the least: and of the others, FB shall be greater than FC, and FC greater than FG.

*

ВА

Join BE, CE, GE: and because two sides of a triangle are greater than the third, therefore BE, EF, are +15 Def. 1. greater than BF: but AE is equal + to BE; 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, each +9 Ax. 1. to each; but the angle BEF is greater + than the angle CEF; therefore the base

* 24. 1.

* 20. 1.

† 5 Ax.

* 23. 1.

+ Constr.

* 4. 1.

*

G

BF is greater than the base CF: for the same reason, CF is greater than GF: again, because GF, FE are greater than EG, and EG is equal to ED; therefore GF, FE are greater than ED: take 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.

*

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 the angle FEH equal to the †15 Def. 1. angle FEG, 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, each to each; and the angle GEF is equal to the angle HEF; therefore the base FG is equal 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 + 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.

+1 Ax.

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 passing 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 straight lines can be drawn from the same point to the circumference, one upon each side of the least 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 part of the circumference AEFC, the greatest shall be DA, which passes through the centre; and the nearer to it. shall be greater than the more remote, viz. DE greater than DF, and DF greater than DC: but of those which fall upon the convex circumference HLKG, the least shall be DG between the point D and the diameter AG; and the nearer to it shall be less than the more remote, viz. DK less than DL, and DL less than DH.

Ꭰ

* 20.1.

Take M the centre of the circle ABC, and join *1.3. ME, MF, MC, MK, ML, MH. And because AM is equal to ME, add MD to each, therefore AD is equal† † 2 Ax. to EM, MD: but EM, MD are greater * 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, each to each: but the angle EMD is greater+ than the angle FMD; therefore the base ED is greater than the base FD. 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 greater than DC. And because MK, KD are greater than MD, and MK is equal + to MG, the remainder KD is greater*

*

HAGEN

M

† 9 Ax.

* 24. 1.

*20.1.

+15 Def. 1.

• 5 Ax.