III. A portion of the circumference, such as FHG, is called

an arc.

The chord or subtense of an arc is the straight line FG, which joins its two extremities.

IV. A segment is the surface, or portion of a circle, included between an arc and its chord.

Note. In all cases, the same chord FG belongs to two arcs, FHG, FEG, and consequently also to two segments: but the smaller one is always meant, unless the contrary is expressed.

V. A sector is the part of the circle included between an arc DE, and the two radii CD, CE, drawn to the extremities of the arc.

VI. A line is said to be inscribed in a circle, when its extremities are in the circumference, as AB.

An inscribed angle is one which, like BAC, has its vertex in the circumference, and is formed by two chords.

An inscribed triangle is one which, like BAC, has its three angular points in the circumference.

And, generally, an inscribed figure is one, of which all the angles have their vertices in the circumference. The circle is said to circumscribe such a figure.

VII. A secant is a line which meets the circumference in two points. AB is a secant.

VIII. A tangent is a line which has but one point in common with the circumference. CD is a tangent.

The point M is called the point of contact.

M

IX. In like manner, two circumferences touch each other when they have but one point in common.

X. A polygon is circumscribed about a circle, when all its sides are tangents to the circumference (see the diagram of Prop. 6. IV.): in the same case, the circle is said to be inscribed in the polygon.

PROPOSITION I. THEOREM.

Every diameter AB divides the circle and its circumference into two equal parts.

For, if the figure AEB be applied to AFB, their common base AB retaining its position, the curve line AEB must fall exactly on the curve line AFB, otherwise there would, in A the one or the other, be points unequally distant from the centre, which is contrary to the definition of a circle.

PROPOSITION II. THEOREM.

Every chord is less than the diameter.

B

For, if the radii AC, CD (see the last figure) be drawn to the extremities of the chord AD, we shall have the straight line AD AC+CD, or AD AB.

Cor. Hence, the greatest line which can be inscribed in a circle is equal to its diameter.

PROPOSITION III. THEOREM.

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

For, if it could meet it in three, those three points would be equally distant from the centre; and hence, there would be three equal straight lines drawn from the same point to the same straight line, which is impossible (Prop. 16. I.).

PROPOSITION IV. THEOREM.

In the same circle, or in equal circles, equal arcs are subtended by equal chords; and, conversely, equal chords subtend equal

arcs.

If the radii AC, EO are equal, and the arcs AMD, ENG; then the chord AD will be equal to the chord A EG.

For, since the diameters AB, EF are equal, the semicircle AMDB may be ap

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plied exactly to the semicircle ENGF, and the curve line AMDB will coincide entirely with the curve line ENGF. But the part AMD is equal to the part ENG (Hyp.); hence the point D will fall on G; hence the chord AD is equal to the chord EG.

Conversely, supposing again the radii AC, EO to be equal, if the chord AD is equal to the chord EG, the arcs AMD, ENG will be equal.

For, if the radii CD, OG be drawn, the triangles ACD, EOG, having all their sides respectively equal, namely, AC=EO, CD-OG, and AD EG, are themselves equal; and, consequently, the angle ACD is equal EOG. Now, placing the semicircle ADB on its equal EGF, since the angles ACD, EOG are equal, it is plain that the radius CD will fall on the radius OG, and the point D on the point G; therefore the arc AMD is equal to the arc ENG.

PROPOSITION V. THEOREM.

In the same circle, or in equal circles, a greater arc is subtended by a greater chord, and conversely, the arcs being always supposed to be less than a semicircumference.

Let the arc AH be greater than AD (see the preceding figure); and draw the chords AD, AH, and the radii CD, CH. The two sides AC, CH of the triangle ACH are equal to the two AC, CD of the triangle ACD, and the angle ACH is greater than ACD; hence (Prop. 10. I.) the third side AH is greater than the third AD; hence the chord, which subtends the greater arc, is the greater.

Conversely, if the chord AH is greater than AD, it will follow, on comparing the same triangles, that the angle ACH is greater than ACD; and hence, that the arc AH is greater than AD.

Scholium. The arcs here treated of are each less than the semicircumference. If they were greater, the reverse property would have place; as the arcs increased, the chords would diminish, and conversely. Thus, the arc AKBD being greater than AKBH, the chord AD of the first is less than the chord AH of the second.

PROPOSITION VI. THEOREM.

The radius CG, at right angles to a chord AB, divides it, and the subtended arc AGB, each into two equal parts.

DRAW the radii CA, CB. These radii considered with regard to the perpendicular CD, are two equal oblique lines; hence (Prop. 16. I.) they lie equally distant from that perpendicular: hence AD is equal to AD.

Again, since AD, DB are equal, CG is A a perpendicular erected from the middle of AB; hence (Prop. 17. I.) every point of this perpendicular must be equally dis

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tant from its two extremities A and B. 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; hence, 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.

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, 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.

PROPOSITION VII. THEOREM.

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

would also be perpendicular to FG (23. I.); and the angle K would be a right angle; but BK, the production 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 (15. I.) is impossible; 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 (17. I.); the same point O, since it lies in the perpendicular FG, is also equally distant from the two points B and C: hence 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 shewn that one circumference can always be made to pass through three given points, not in the same straight line: we assert farther, that but one can be made to

pass.

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 (17. I.), for then it would be unequally distant from A and B; no more 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; hence there is but one circumference which can pass through three given points.

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.

Two equal chords are equally distant from the centre; and two unequal chords, the less is farther 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, D DCG, the hypotenuses CA, CD are equal; and the side AF, the half of AB, is equal to the side DG, the half of DE: hence the triangles are equal (18. I.), and CF is equal to CG; hence (first) the two equal chords AB, DE, are equally distant from the centre.

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