ties A and B. (16. 1.) Now, G is one of those points; therefore AG, BG are equal. But, if the chord AG is equal to the chord GB, the arc AG will be equal to the arc GB; (4.2;) therefore the radius CG, at right-angles to the chord AB, divides the arc subtended by that chord into two equal parts at the point G. Hence,

Any radius which is at right-angles to a chord, bisects both the chord, and the subtended arc.

Scholium. The centre C, the middle point D of the chord AB, and the middle point G of the arc subtended by this chord, are three points situated in the same line perpendicular to the chord. But two points are sufficient to determine the position of a straight line; hence every straight line which passes through two of the points just mentioned, will necessarily pass through the third, and be perpendicular to the chord.

It follows, likewise, that the perpendicular, raised from the middle of a chord, passes through the centre of the given circle, and through the middle of the arc subtended by that chord.

For this perpendicular is identical with the one let fall from the centre on the same chord, since both of them pass through the middle of the chord.

Through three given points A, B, C, not in the same straight line, one circumference may always be made io pass, and only

they were parallel, the line AB, which is perpendicular to DE, would also be perpendicular to FG; (23. 1. Cor.;) and the angle K would be a right-angle; but BK, the extension of BD, is different from BF, because the three points A, B, C, are not in the same straight line; hence there would be two perpendiculars, BF, BK, let fall from the same point on the same straight line, which is impossible; (14. 1;) hence DE, FG will always meet in some point O.

And moreover, this point O, since it lies in the perpendicular DE, is equally distant from the two points A and B; (16. 1;) the same point O, since it lies in the perpendicular FG, is also equally distant from the two points B and C : therefore the three distances OA, OB, OC, are equal; hence the circumference described from the centre O, with the radius OB, will pass through the three given points A, B, C.

We have now shown that one circumference can always be made to pass through three given points, not in the same straight line: we assert farther, that only one can be made to pass through them.

For, if there were a second circumference passing through the three given points A, B, C, its centre could not be out of the line DE, for then it would be unequally distant from A and B; (16. 1;) nor could it be out of the line FG, for a like reason; therefore, it would be in both the lines DE, FG. But two straight lines cannot cut each other in more than one point; consequently there is but one circumference which can pass through three given points. Hence,

One circumference, and only one, may always be made to pass through three given points, not in the same straight line.

Cor. Two circumferences cannot meet in more than two points; for, if they have three common points, they must have the same centre, and form one and the same circumference.

PROPOSITION VIII. THEOREM.

Equal chords are equally distant from the centre of a circle; and of unequal chords, the less is the farthest from the centre.

First. Suppose the chord AB-DE. Bisect those chords by the perpendiculars CF, CG, and draw the radii CA, CD.

In the right-angled triangles CAF, DCG, the hypothenuses CA, CD are equal; and the side AF, the half of AB, is

equal to the side DG, the half

E

M

D

C

H

F

N

K

of DE hence the triangles are equal, and CF is equal to CG; (17. 1;) hence the two equal chords AB, DE, are equally distant from the centre.

Second. Let the chord AH be greater than DE. The arc AKH will be greater than DME: (5. 2 :) cut off from the former, a part ANB=DME the latter; draw the chord AB, and let fall CF perpendicular to this chord, and CI perpendicular to AH. It is evident that CF is greater than CO, and CO than CI; (15. 1;) therefore, CF is still greater than CI. But CF is equal to CG, because the chords AB, DE, are equal; hence we have CG>CI; therefore of two unequal chords, the less is the farther from the centre. Hence, Two equal chords, &c.

PROPOSITION IX. THEOREM.

The straight line BD, perpendicular to the radius CA, and passing through its extremity, is a tangent to the circumference.

For, every oblique line CE, is longer than the perpendicular CA ; (15. 1;) hence the point E is without the circle; hence BD

has no point but A common to it with the circumference; therefore

.H

E

B

-D

BD is a tangent. (Def. 11. 2.) Hence,

Every straight line perpendicular to a radius, and passing through its extremity, is a tangent to the circumference.

Scholium. From a given point A, only one tangent AD can be drawn to the circumference; for if another FH, could be drawn, it would not be perpendicular to the radius CA; (14. 1. Sch.;) hence, in reference to this new tangent FH, the radius AC would be an oblique line, and the perpendicular CG let fall from the centre upon this tangent FH, would

be shorter than CA; hence this supposed tangent FH, would enter the circle, and be a secant. (Def. 10. 2.)

PROPOSITION X. THEOREM.

Two parallels AB, DE, intercept equal arcs MN, PQ, on the circumference.

There may be three cases. First. If the two parallels are secants, draw the radius CH perpendicular to the chord MP. It will, at the same time, be perpendicular to NQ; (23. 1. Cor.;) therefore, the point H, will be at once the middle of the arc MHP, and

of the arc NHQ; (6. 2;) hence,

H

D

E

N

G

Q

A

B

M

F

C

we shall have the arc MH-HP, and the arc NH=HQ; and therefore MH-NH=HP-HQ; in other words, MN=PQ.

Second. When, of the two parallels AB, DE, one is a secant, and the other a tangent, draw the radius CH to the point of contact H; it will be perpendicular to the tangent DE, (9. 2,) and also to its parallel MP. But, since CH is perpendicular to the chord MP,

H

D

E

A

B

M

F

c+

I

-L

K

the point H must be the middle of

the arc MHP; (6.2;) therefore the arcs MH, HP, included between the parallels AB, DE, are equal.

Third. If the two parallels DE, IL are tangents, the one at H, the other at K, draw the parallel secant AB; and, from what has just been shown, we shall have MH=HP,