b 3.3.

Book III. does not pass through the centre, it must cut 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 must cut it at right angles; wherefore FEB is a right angle. But 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 each other. Wherefore, if in a circle, &c. Q. E. D.

PROP. V. THEOR.

IF two circles cut each other they cannot have the same centre.

Let the two circles ABC, CDG cut each other 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.
Because E is the centre of
the circle ABC, CE is equal
to EF; and because E is the
centre of the circle CDG, CE A
is equal to EG; 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.

D

F

E

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

PROP. VI. THEOR.

IF two circles touch each other internally they cannot have the same centre.

Let the two circles ABC, CDE touch each other 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 them in E and B; then, because F is the centre of the circle ABC, CF is equal to FB; also, because F is the centre

of the circle CDE, CF is equal to FE; therefore FE is equal to FB, the less to the greater, which is impossible. Wherefore F is not

F

D

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 all other lines 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 each other, 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 dentre

Book III.

20. 1.

A

B

Book III. 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 diameter AD, is the least; and of the other lines FB is greater than FC, and FC than FG. Join BE, CE, GE; then, BE, EF are greater than BFa. But AE is equal to EB; therefore AE, EF, that is, AF, is greater than F. 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 base BF is greater than the base FC. For the same reason CF is greater than GF. Again, because GF, FE are greater than EG, and EG is equal to ED, GF,

b 24. 1.

c 23. 1.

d 4. 1.

E

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 drawn 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 angle GEF, and join FII; 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 to the angle HEF; therefore the base FG is equal to the base FH. But, beside FH, no straight line can be drawn from F to the circumference equal to FG; for, if there can, let it be FK; then, 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 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.

Book III.

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 nearer to it is always greater than the more remote, viz. DE than DF, and DF than DC; 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.

Take M the centre of the circle ABC, and join ME, MF, a 1. 3. MC, MK, ML, MH. 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 than ED; therefore also AD is b 20. 1. greater than ED. Again, because ME is equal to MF, and

Book III. MD common to the triangles

c 24. 1.

d 4. Ax.

e 21. 1.

f 23. 1.

g 4. 1.

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 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

GB

BN

M

E

A

Because MK, KD are greaterb than MD, and MK is equal to MG, the remainder KD is greater than the remainder GD, that is, GD is less than KD. 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 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 DC is the least, and DK less than DL, and DL than DH.

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, makef the angle DMB equal to the angle DMK, and join DB; then 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, therefore the base DK is equals to the base DB. But, beside DB, no straight fine 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.