PROP. II. THEOR.

A straight line (AB) joining two points in the circumference of a circle (AGB), lies wholly within the circle.

Find the centre C (iii. Prop. 1); draw CA, CB; take any point D in the straight line AB, and draw CD,

meeting the circumference in G.

Then, since CA = CB, ▲ CAB

= CBA

(i. Prop. 5): but CDA is greater than

CBA

D

B

G

(i. Prop. 16), and is therefore also greater than /CAB or /CAD, and therefore CĂ subtending the greater angle in the triangle CAD is greater than CD subtending the less (i. Prop. 19); but CA = CG, and therefore CD is less than the ray CG, and consequently the point D lies within the circle: and the same may be proved of every other point in the line AB; and therefore the straight line joining two points in the circumference, must be wholly within the circle.

PROP. III. THEOR.

If a straight line (FE) drawn through the centre of a circle (AFB) bisects a chord (AB) which does not pass through the centre, it is perpendicular to it; or, if perpendicular to it, it bisects it.

Find the centre C, and join CA, CB. Then, since FE bisects AB in D, DA=DB; and CD is common to the two triangles CDA, CDB; and also CA=CB; therefore CDA =/ CDB, and FE is perpendicular to AB (i. Def. 7).

Again, let FE be perpendicular to AB. Then, because in the triangles ACD, BCD, the side CACB; and also / CAD = ≤ CBD (i. Prop. 5),

B

and CDA =/ CDB (Hyp.), the sides between the equal angles are also equal (i. Prop. 26), or AD = DB, and therefore FE bisects AB.

PROP. IV. THEOR.

Two chords (AB, CD) in a circle, intersecting in a point (E), which is not the centre, cannot bisect each other.

If one of the chords passes through the centre, it is evident that it is not bisected by the other which does not pass through the centre. But if neither of them passes through the centre, then find the centre F, and join FE. And if it be supposed that the chords AB, CD are both bisected in the point E, then

since FE drawn from the centre bisects AB in E, it is also perpendicular to it (iii. Prop. 3), and FEB is a right angle. And in like manner, since FE bisects CD, FED is a right angle; and therefore ▲ FED=▲ FEB, a part equal to the whole, which is absurd. Therefore the chords AB and CD are not both bisected in the point E.

PROP. V. THEOR.

If two circles (AEB, AFB) intersect, they have not the

same centre.

E

For suppose it possible that the two circles intersecting in the point A have the common centre C. Join CA, and also draw CE, meeting the circumferences of the circles where they do not intersect in D and E. Then, because C is the centre of the circle ADB, CD=CA (i. Def. 12); and in like manner, be

cause C is the centre of the circle AEB, CA=CE; therefore CDCE; a part equal to the whole, which is absurd. Therefore C is not the centre of both circles; and the same may be demonstrated of every other point. Therefore circles intersecting cannot have the same centre.

PROP. VI. THEOR.

If two circles (FBA, FDE) touch internally, they have not the same centre.

1

F

D

For suppose it possible that the two circles, the one touching the other internally at the point F, have the common centre C. Join CF, and also draw CB, meeting the circumferences of the circles where they do not touch, in D and B. Then, because C is the centre of the circle FDE, CD= CF (i. Def. 12), and again, because C is the centre of the circle FBA, CF=CB; and therefore CD=CB, a part equal to the whole, which is absurd. Therefore C is not the centre of both circles, and the same may be demonstrated of every other point; and consequently circles which touch internally cannot have the same centre.

A

If from a point (A) within a circle (BEH), but which is not its centre, lines (AB, AD, AE) are drawn to the circumference; the greatest of those lines is that (AB) which passes through the centre (c), and the least is the remaining part (AH) of the diameter. Of the others, that (AD) which is nearer to the line passing through the centre, is greater than that (AE) which is more remote; the two lines (AE, AL) which make equal angles with that passing through the centre, on opposite sides of it, are equal to each other; and there cannot be drawn a third line equal to them, from the same point to the circumference.

B

Join CD and CE; then because CB=CD (i. Def. 12), the line AB, passing through the centre is equal to AC and CD together; but AC and CD together are greater than AD (i. Prop. 20), therefore AB is greater than AD; and in like manner it may be shown to be greater than AE or any other line drawn from the point A to the circumference. Again, since

H

CE=CH, but CA and AE together are greater than CE,

they are also greater than CH; take CA from both, and AE remains greater than AH (Ax. 5). And in like manner it may be shown that AH is less than any other line drawn from the point A to the circumference. And again, since in the triangles ACD, ACE, the side CD=CE, CA is common to both, and ACD is greater than ACE, the base AD is greater than AE (i. Prop. 24); and AD may in like manner be proved greater than any other line drawn from A to the circumference more remote from AB. Also, the line AL, making with the line passing through the centre, BAL=/ BAE, is equal to AE. For if it be supposed that L AL is not equal to AE, make AG equal r to AE, and draw CG, meeting the circumference in F; then, because in the triangles

A

B

H

D

E

AGC, AEC, the side AG=AE, AC is common, and CAG =/ CAE, the base CG=CE (i. Prop. 4); but CE=CF, and therefore CG=CF, a part equal to the whole, which is absurd. Therefore AG is not equal to AE, and the same may be demonstrated of any other segment of AL, which is consequently itself equal to AE. Also, any other line drawn from the point A to the circumference, must either pass through the centre, and consequently be greater than AE or AL, or else it must lie at the same side with one of them, of the line passing through the centre; and the two lines so lying at the same side of the line passing through the centre, must be at unequal distances from it, and therefore be unequal. There cannot therefore be drawn a third line equal to those making equal angles with the line passing through the centre.

PROP. VIII. THEOR.

If from a point (A) without a circle, straight lines (AB, AD, AE) are drawn to the circumference; of those falling upon the concave circumference the greatest is that (AB) which passes through the centre, and the line which is nearer the greatest is greater than that which is more remote: of those falling on the convex circumference the least is that (AB) which being produced would pass through the centre, and the line which is nearer to the least is less than that which is

more remote; and lines (AE, AG), whether falling on the concave or convex circumference, which make equal angles with that passing through the centre, are equal to each other; and no third line can be drawn equal to them from the same point to the circumference.

Then, of the lines

D B

E

L

G

Find the centre C, and join CD, CE. falling on the concave circumference, AB, which passes through the centre, is greatest; for since CB=CD, if AC be added to both, AB=AC + CD; but AC and CD together are greater than AD (i. Prop. 20), therefore AB, which passes through the centre, is also greater than AD; and in like manner it may be shown to be greater than any other line drawn from the point A to the concave circumference and not passing through the centre. Again, since in the two triangles ACD, ACE the side CD = CE, AC is common to both, and ACD is greater than ACE, the base AD is greater than AE (i. Prop. 24); and in like manner AD may be shown to be greater than any other line more remote from AB.

D

Ᏼ

But of the lines which fall on the convex circumference, AB, which produced would pass through the centre, is the least. For, since CD and DA are together greater than CA (i. Prop. 20), and CB = CD, if these be taken away, the remainder AD will be greater than AB (Ax. 5). And again, since CD and DA are together less than CE and EA (i. Prop. 21), but CD=CE, if these be taken away, AD, which is nearer the least line, will be less than AE, which is more remote from it.

Also the lines making equal angles with that which passes through the centre are equal. For if it be supposed that AG and AE making equal angles with AB are unequal, and that AG is the greater; cut off from it AF equal to E, and draw CF meeting the circumference in L. Then, since the triangles CAF, CAE have AF = AE, AC common, and ▲ CAF = CAE, they have their bases equal, and CF CE (i. Prop. 4); but CE = CL; therefore CF CL, a part equal to the whole, which is absurd. Therefore AE is not equal to AF, nor to any other part of AG, which is therefore not greater than AE. And in like manner it may be shown

=

E

=