Then, if the proposition is not true, let us suppose that the angle ACB is to the angle FOH, or its equal ACD, as the arc AB is to an arc AO, greater than FH, or its equal AD; whence,

DIOB I

angle ACB : angle ACD :: arc AB

Conceive the arc each less than DO: division between D

arc AO.

АВ to be divided into equal parts, there will be at least one point of and ; let I be that point; and draw CI Then the arcs AB, AI, will be commensurable, and we shall have (P. XVI.),

angle ACB : angle ACI :: arc AB : arc AI

Comparing the two proportions, we see that the antecedents are the same in both: hence, the consequents are proportional (B. II., P. IV., C.); hence,

angle ACD : angle ACI

arc 40: arc AI

But, AO is greater than AI: hence, if this proportion is true, the angle ACD must be greater than the angle ACI. On the contrary, it is less: hence, the fourth term of the proportion cannot be greater than AD.

In a similar manner, it may be shown that the fourth term cannot be less than AD: hence, it must be equal to AD; therefore, we have,

angle ACB : angle ACD :: arc AB

which was to be proved.

arc AD

Cor. 1. The intercepted arcs are proportional to the cor

responding angles at the centre, as may be shown by changing the order of the couplets in the preceding proportion.

Cor. 2. In equal circles, angles at the centre are proportional to their intercepted arcs; and the reverse, whether they are commensurable or incommensurable.

Cor 3. In equal circles, sectors are proportional to their angles, and also to their arcs.

Scholium. Since the intercepted arcs are proportional to the corresponding angles at the centre, the arcs may be taken as the measures of the angles. That is, if a circumference be described from the vertex of any angle, as a centre, and with a fixed radius, the arc intercepted between the sides of the angle may be taken as the measure of the angle. In Geometry, the right angle which is measured by a quarter of a circumference, or a quadrant, is taken as a unit. If, therefore, any angle be measured by one-half or two-thirds of a quadrant, it will be equal to one-half or two-thirds of a right angle.

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

There may be three cases: the centre of the circle may

lie on one of the sides of the angle; it

may lie within the angle; or, it may die without the angle.

1o. Let EAD be an inscribed angle, one of whose sides AE passes through the centre: then will it be measured by half of the arc DE

E

For, draw the radius CD. The external angle DCE, of the triangle DCA, is equal to the sum of the opposite interior angles CAD and CDA (B. L., P. XXV., C. 6). But, the triangle DCA being isosceles,

the angles D and A are equal;

therefore, the angle DCE is double the angle DAE. Because DCE is at the centre, it is measured by the arc DE (P. XVII., S.): hence, the, angle DAE is measured by half of the arc DE; which was to be proved.

B

E

2°. Let DAB be an inscribed angle, and let the centre lie within it then will the angle be measured by half of the arc BED.

Then, from what has just measured by half of DE, EB: hence, BAD, which

For, draw the diameter AE. been proved, the angle DAE is and the angle EAB by half of is the sum of EAB and DAE, is measured by half of the sum of DE and EB, or by half of BED; which was to be proved.

3°.

Let BAD be an inscribed angle, and let the centre lie without it: then will it be measured by half of the arc arc BD.

For, draw the diameter AE. Then, from what precedes, the angle DAE is measured by half of DE, and the angle BAE by half of BE: hence, BAD, which is the difference of BAE and DAE, is measured by half of the difference of BE

and DE, or by

B

D

half of the arc BD; which was to be proved.

obtuse; for it is measured by half 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 DAB is measured by half the arc DCB, the angle DCB by half the arc

:

D

B

DAB hence, the two angles, taken together, are measured by half the circumference: hence, their sum is equal to two right angles.

Any angle formed by two chords, which intersect, is measured by half the sum of the included arcs.

Let DEB be an angle formed by the intersection of the chords AB and CD:

then will it be measured by

half the sum of the arcs AC and DB.

XVIII.); therefore, DEB is measured

by half of FDB; that is, by half the sum of FD and DB, or by half the sum of AC and DB; which was to be proved.

PROPOSITION XX. THEOREM.

The angle formed by two secants, is measured by half the difference of the included arcs.

Let AB, AC, be two secants: then will the angle BAC be measured by half the difference of the arcs BC and DF

Draw DE parallel to AC: the arc EC will be equal to DF (P. X.), and the angle BDE equal to the angle BAC (B. I., P. XX., C. 3.). But BDE is measured by half the arc BE (P. XVIII.): hence, BAC is also measured by half the arc BE;

that is, by half the difference of BC

B

and EC, or by half the difference of BC and DF; which was to be proved.