gles in one equal to two angles in the other, and the fide EF which Book III. is oppofite to one of the equal angles in each, is common to both; therefore the other fides are equal, AF therefore is equal to FB. e. 26. 1. Wherefore if a ftraight line, &c. Q. E. D.

F in a circle two straight lines cut one another which do not both pass thro' the center, they do not bifect each the 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 thro' the center. AC, BD do not bifect one another.

For, if it is poffible, let AE be equal to EC, and BE to ED. if one of the lines pass thro' the center, it is plain that it cannot be bifected by the other which does not pass thro' the center. but if neither of them pass thro' the center, take a F the eenter of the circle, and join EF. and because FE a straight line thro' the center, bifects another AC which does A not pafs thro' the center, it fhall cut it at right bangles; wherefore FEA is a right angle. again, because the straight

F

BE

line FE bisects the straight line BD which does not pass thro' the
center, it shall cut it at right angles; wherefore FEB is a right
angle. and FEA was fhewn to be a right angle; therefore FEA is
equal to the angle FEB, the lefs to the greater, which is impof-
fible. therefore AC, BD do not bifect one another.
if in a circle, &c. Q. E. D.

Wherefore

PROP. V. THEOR.

two circles cut one another, they shall not have
the fame center.

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

a. 1. 3.

b. 3. 3.

Book III.

For, if it be poffible, let E be their center; join EC, and draw any straight line EFG meeting

them in F and G. and because E
is the center of the circle ABC,
CE is equal to EF. again, because
E is the center of the circle CDG,
CE is equal to EG. but CE was
fhewn to be equal to EF; there-
fore EF is equal to EG, the lefs
to the greater, which is impoffible.
therefore E is not the center of

A

E

B

the circles ABC, CDG. Wherefore if two circles, &c. Q. E. D.

PROP. VI. THEOR.

two circles touch one another internally, they fhall not have the fame center.

Let the two circles ABC, CDE touch one another internally in the point C. they have not the fame center.

For if they can, let it be F; join FC, and draw any straight

line FEB meeting them in E and B.

and because F is the center of the
circle ABC, CF is equal to FB. alfo
because F is the center of the circle
CDE, CF is equal to FE. and CF
was fhewn equal to FB; therefore
FE is equal to FB, the lefs to the
greater, which is impoffible. where-
fore F is not the center of the cir-

A

B

F

E

D

cles ABC, CDE. Therefore if two circles, &c. Q. E. D.

IF

PROP. VII. THEOR.

F any point be taken in the diameter of a circle which is not the center, of all the ftraight lines which can be drawn from it to the circumference, the greatest is that in which the center is, and the other part of that diameter is the leaft; and of any others, that which is nearer to the line which paffes thro' the center is always greater than one more remote. and from the fame point there can be drawn only two ftraight lines that are equal to one another, one upon each fide 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 center. let the center 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 leaft; and of the others, FB is greater than FC, and FC than FG.

Book III.

Join BE, CE, GE; and because two fides of a triangle are greater than the third, BE, EF are greater than BF; but AE is a. 2ẹ. 1.

a

equal to EB, therefore AE, EF, that is AF, is greater than BF. again, becaufe BE is equal to CE, and FE common to the triangles BEF, CEF; the two fides 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 fame reafon, CF is greater than GF. again, becaufe GF, FE

a

b

are greater 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 than the remainder FD. therefore FA is the greateft, and FD the least of all the ftraight lines from F to the circumference; and BF is greater than CF, and CF than GF.

Also there can be drawn only two equal ftraight lines from the point F to the circumference, one upon each fide of the shortest line

E

C. 23. I.

C

B

Book III. FD. at the point E in the straight line EF, make 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 fides GE, EF are equal to the two HE, EF; and the angle GEFis equal to the angle HEF, therefore the base FG is equal to the bafe FH. but befides FH no other ftraight line can be drawn from F to

d. 4. 1.

d

the circumference equal to FG. for

E

D

H

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 paffes thro' the center is equal to one which is more remote; which is impoffible. Therefore if any point be taken, &c. Q. E. D.

IF

any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one paffes thro' the center; of those which fall upon the concave circumference the greateft is that which paffes thro' the center; and of the rest, that which is nearer to that thro' the center is always greater than the more remote. but of thofe which fall upon the convex circumference, the leaft is that between the point without the circle, and the diameter; and of the reft, that which is nearer to the least is always lefs than the more remote. and only two equal ftraight 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 ftraight lines DA, DE, DF, DC be drawn to the circumference, whereof DA paffes thro' the center. of those which fall upon the concave part of the circumference AEFC, the greateft is AD which paffes thro' the center; 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

leaft is DG between the point D and the diameter AG; and the Book III. nearer to it is always lefs than the more remote, viz. DK than DL, and DL than DH.

Take a M the center of the circle ABC, and join ME, MF, MC, a. 1. 3 MK, ML, MH. and because AM is equal to ME, add MD to each, therefore AD is equal to EM, MD; but EM, MD are greater b b. 20. In than ED, therefore alfo AD is greater than F.D. again, because ME is equal to MF, and MD common to the triangles EMD, FMD; EM, MD are equal to FM,

C

d

H

D.

C. 24. I

GB

N

d. 4. As.

M

1

E

A

e. 21. I.

MD; but the angle EMD is greater than the angle FMD, therefore the bafe ED is greater than the base FD. in like manner it may be fhewn that FD is greater than CD. therefore DA is the greatest; and DE greater than DF, and DF than DC. and because MK, KD are greater b than MD, and MK is equal to MG, the remainder KD is greater than the remainder GD, that is, GD is C lefs than KD. and because MK, DK are drawn to the point K within the F triangle MLD from M, D the extremities of its fide MD; MK, KD are lefs than ML, LD, whereof MK is equal to ML, therefore the remainder DK is lefs than the remainder DL. in like manner it may be fhewn that DL is less than DH. therefore DG is the least, and DK less than DL, and DL than DH. Also there can be drawn only two equal ftraight 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 MK is equal to MB, and MD common to the triangles KMD, BMD, the two fides KM, MD are equal to the two BM, MD; and the angle KMD is equal to the angle BMD, therefore the base DK is equal f to the f. 4. &. base DB. but befides DB there can be no straight line drawn from D to the circumference equal to DK. for if there can, let it be DN; and becaufe DK is equal to DN, and alfo to DB, therefore DB is equal to DN, that is the nearer to the leaft equal to the more remote, which is impoffible. If therefore any point, &c. Q. E. D.