MK KP; and hence the whole arc HMK=HPK. It is also evident that each of these arcs is a semicircumferHence,

ence.

Any two parallels intercept equal arcs on the circumference..

In the same circle, or in equal circles, the equal angles ACB, DCE, having their vertices in the centre, intercept equal arcs AB, DE on the circumference; and conversely, if the arcs AB, DE are equal, the angles ACB, DCE will also be equal. First. Since the angles

ACB, DCE are equal, they may be placed upon each other; and since their sides are equal, the point A will evidently fall on D, and the point B on E. But, in

B D

that case, the arc AB must also fall on the arc DE; for if they did not exactly coincide, there would, in the one or the other, be points unequally distant from the centre; which is impossible; (Def. 1. 2;) hence the arc AB is equal to DE.

Second. If we suppose AB=DE, the angle ACB will be equal to DCE. For if those angles are not equal, let ACB be the greater, and let ACI be taken equal to DCE. From what has just been shown, we shall have AI-DE: but, by hypothesis, AB is equal to DE; hence AI must be equal to AB, a part to the whole, which is absurd; (Ax. 9;) consequently, the angle ACB is equal to DCE. Hence,

In the same circle, or in equal circles, equal angles having their vertices in the centre, intercept equal arcs on the circumference; and, conversely, if the arcs intercepted by radii, are equal, the angles contained by those radii, are equal.

In the same circle, or in equal circles, two angles at the centre ACB, DCE are to each other as the intercepted arcs AB, DE, by which the angles are subtended; that is, the angle ACB is to the angle DCE, as the arc AB to the arc DE.

Suppose, e. g., that the angles ACB, DCE are to each other as 6 is to 3; or, which is the same thing, suppose that the angle M, which may serve as a common measure, is contained six times in the angle ACB and three times in DCE. The six partial angles ACm, mCn, nCp, &c., into which ACB is divided, being each equal to any of the three partial angles into which DCE is divided; each of the partial arcs Am, mn, np, &c., will consequently be equal to each of the partial arcs Ax, xy, &c. (11.2.) Therefore the whole arc AB will be to the whole arc DE as 6 is to 3. But the same reasoning would evidently apply, if in place of 6 and 3, any numbers whatever were employed; hence, the angles ACB, DCE are to each other as the arcs AB, DE. Therefore,

All angles at the centre of the same circle, or equal circles, are to each other, as the arcs by which they are subtended.

Scholium 1. Conversely, the arcs AB, DE are to each other as the angles, ACB, DCE: i. e. the angle ACB is to the angle DCE as the arc AB is to the arc DE. For the partial arcs, Am, mn, &c., and Dx, xy, &c., being equal, the partial angles ACm, mCn, &c. and DC x, xCy, &c. will also be equal.

Scholium 2. All that has been demonstrated in the last two propositions, concerning the comparison of angles with arcs, holds true equally, if applied to the comparison of sectors with arcs; for sectors are not only equal when their angles are so, but in all respects are proportional to their angles; therefore two sectors ACB, DCE, taken in the same circle, or in equal circles, are to each other as the arcs AB, DE, the bases of those sectors. Hence it is evident that the arcs of the circle, which serve as a measure of the different angles, may also serve as a measure of the different sectors, in the same circle, or in equal circles.

Cor. Since the angle at the centre of a circle, and the arc intercepted by its sides, have such a connection, that if the one be augmented or diminished in any ratio, the other will be augmented or diminished in the same ratio, we are authorised to establish the one of those magnitudes as the measure of the other; and we shall henceforth assume the arc AB as the measure of the angle ACB. It is only required that, in the comparison of angles with each other, the arcs which serve to measure them, be described with equal radii, as the foregoing propositions imply.

The inscribed angle BAD, is measured by half the arc BD, included between its sides.

First. Suppose that the centre of the circle lies within the angle BAD. Draw the diameter AE, and the radii CB, CD.

The angle BCE, being exterior to the triangle ABC, is equal to the sum of the two interior angles CAB, ABC: (28.1. Cor. 6:) but the triangle

B

E

BAC being isosceles, the angle CAB is equal to ABC; hence the angle BCE is double of BAC. Since BCE lies at the centre, it is measured by the arc BE; hence BAC will be measured by the half of BE. (12. 2.) For a like reason, the angle CAD will be measured by the half of ED; hence BAC+CAD, or BAD will be measured by the half of BE+ED, or of BD.

Second. Suppose that C the centre lies without the angle BAD. Then, drawing the diameter AE, the angle BAE will be measured by the half of BE; the angle DAE by the half of DE: hence their difference BAD will be measured by the half of BE minus the half of ED, or by the half of BD. Therefore,

DE

C

Every inscribed angle is measured by the half of the arc included between its sides.

Cor. 1. All the angles BAC, BDC, inscribed in the same segment are equal; because they are all measured by the half of the same arc BOC.

Cor. 2. Every angle BAD, inscribed in a semicircle, is a right-angle; because it is measured by half the semi- B circumference BOD; that is, by the

fourth part of the whole circumference.

D

E

D

Cor. 3. Every angle BAC, (see the diagram of Cor. 1,) inscribed in a segment greater than a semicircle, is an acute

angle; for it is measured by the half of the arc BOC, less than a semi circumference.

And every angle BOC, inscribed in a segment less than a semicircle, is an obtuse angle; for it is measured by the half of the arc BAC, greater than a semi circumference.

Cor. 4. The opposite angles A and C, of an inscribed quadrilateral ABCD, are together equal to two right-angles: for the angle BAD is measured by half the arc BCD, and the angle BCD is measured by half the arc BAD; hence

B

the two angles BAD, BCD, taken together, are measured by the half of the circumference; therefore their sum is equal to two right-angles.

The angle BAC, formed by a tangent and a chord, is measured by the half of the arc AMDC, included between its sides.

From A, the point of contact, draw the diameter AD. The angle BAD is a right-angle, (9.2,) and is

measured by half the circumference AMD; the angle DAC is measured by the half of DC: hence BAD+ DAC, or BAC is measured by the half of AMD plus the half of DC, or by half the whole arc AMDC.

It

M

B

D

E

may be shown in the same manner, that the angle CAE is measured by half the arc AC, included between its sides. Hence,

Every angle formed by a tangent and a chord, is measured by half the arc included between its sides.