Take any two chords AB, CD, which are not parallel. Bisect AB and CD at right angles by the straight lines EO, FO

meeting at O,

I. Prob. 4.

then shall O be the centre required.

For, because EO bisects the chord AB at right angles, therefore the centre is in EO;

III. 10.

and, because FO bisects the chord CD at right angles,

therefore the centre is in FO;

III. 10.

therefore O, the only point common to EO and FO, is the centre of the circle.

Q.E.F.

THEOR. 11. All points in a chord of a circle between the extremities of the chord lie within the circumference; and points in the chord produced either way lie without the circumference.

Let ABC be a circle, DBCE a straight line through B and C, any two points on the circumference:

B

then shall all points in BC between B and C lie within the cir cumference; and all points in BD or CE, except B and C, shall lie without the circumference.

Let O be the centre of the circle, F the middle point of BC, G any point between F and B, H any point in BD; join OF, OG, OB, OH.

Because F is the middle point of the chord BC,

Again, because the angle GOF is less than the angle BOF, therefore OG is less than OB,

therefore G is within the circumference;

I. 15.

III. 1, Cor.

and in like manner any point between F and C is within the circumference

Also, because the angle HOF is greater than the angle BOF, therefore OH is greater than OB;

therefore H is without the circumference;

I. 15.

III. 1, Cor.

and in like manner any point in CE, except C, is without the circumference.

Q.E.D.

COR. A straight line cannot meet the circumference of a circle in more than two points.

DEF. 8. A secant is a straight line of unlimited length which meets the circumference of a circle in two points.

THEOR. 12. There is one circle, and only one, whose circumference passes through three given points not in the same straight line.

line:

Let A, B, C be three given points not in the same straight

B

there is one circle, and only one, whose circumference passes through A, B, C.

Take O the point equidistant from A, B, and C; Int. of Leci, i join OA, OB, OC.

Then, because OA, OB, OC are all equal,

therefore the circumference of a circle drawn with centre O and radius OA will pass through A, B, C.

Again, because there is only one point equidistant from A, B, C, Int. of Loci, i.

and because circles with a common centre and common radius coincide wholly, therefore there is only one circle whose circumference passes through A, B, and C.

Q.E.D.

COR. I.

Two circles whose circumferences have three

points in common coincide wholly.

COR. 2.

The circumferences of two circles which do not coincide cannot meet in more than two points.

COR. 3. If from any point within a circle more than two straight lines drawn to the circumference are equaɔ, that point is the centre.

Ex. 27. Prove that two circles cannot have a common arc.

THEOR. 13. In the same circle, or in equal circles, equal chords are equally distant from the centre; and of two unequal chords the greater is nearer to the centre than the less.

In the equal circles DEF, HKL, whose centres are A and B, let the chord DE be equal to the chord HK, and let AM, BN be perpendicular to DE and HK:

then shall DE and HK be equally distant from the centres, that is, AM shall be equal to BN.

Join AD, BH.

Because AM is drawn from the centre A perpendicular to

and the angles AMD, BNH are right angles;

therefore the side AM is equal to the side BN;

I. 20, Cor. 1.

that is, DE and HK are equally distant from the centres,

Next let the chord DE be greater than the chord HK: then shall AM be less than BN.