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. I., 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.

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

For, draw the diameter AE. Then, from what has just been proved, the angle DAE is measured by half of DE, and the angle EAB by half of EB: hence, BAD, which is the sum of EAB and DAE, is measured by half of the sum of DE and EB, or by half of BED; BED; which was to be proved.

3°. Let BAD be an inscribed angle, and let the centre lie without it: then it is measured by half of the 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 half of the arc BD; which was to be proved.

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

B

Cor. 2. Any angle BAD, inscribed in a semicircle, is a right angle; because it is measured by half the semi-circumference BOD, or by a quadrant (P. XVII., S.).

Cor. 3. Any angle BAC, inscribed in a segment greater than a semicircle, is acute; for it is measured by half the arc BOC, less than a semi-circumference.

Any angle BOC, inscribed in a segment less than a semicircle, is obtuse;

B

B

for it is measured by half the arc BAC, greater than a semi-circumference.

D

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 DAB: hence, the two angles,

Ө

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 it is measured by half the sum of the arcs AC and DB.

C

E

For, draw AD: then, the angle DEB, being an exterior angle of the triangle DEA, is equal to the sum of the angles EDA and EAD (B. I., P. XXV., C. 6). But, the angle EDA is measured by half the arc AC, and EAD by half the arc DB (P. XVIII.): hence, the angle DEB is measured by half the sum of the arcs AC and DB; which was to be proved.

PROPOSITION XX. THEOREM.

The angle formed by two secants, intersecting without the circumference, is measured by half the difference of the included arcs.

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

B

Draw DE parallel to AC: the arc EC is equal to DF (P. X.), and the angle BDE 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 and EC, or by half the difference of BC and DF; which was to be proved.

E

An angle formed by a tangent and a chord meeting it at the point of contact, is measured by half the included

arc.

Let BE be tangent to the circle AMC, and let AC be a chord drawn from the point of contact A: then BAC is measured by half of the arc AMC.

For, draw the diameter AD. The angle BAD is a right angle (P. IX.), and is measured by half the semi-circumference AMD (P. XVII., S.); the angle DAC is measured by half of the arc DC (P. XVIII.): hence, the angle BAC, which

M

B

E

is equal to the sum of the angles BAD and DAC, is measured by half the sum of the arcs AMD and DC, or by half of the arc AMC; which was to be proved.

The angle CAE, which is the difference of DAE and DAC, is measured by half the difference of the arcs DCA and DC, or by half the arc CA.

PRACTICAL APPLICATIONS.

PROBLEM I

To bisect a given straight line.

Let AB be a given straight line. From A and B, as centres, with a radius greater than one half of AB, describe arcs intersecting at E and F: join E and F, by the straight line EF. Then EF bisects the given line AB. For,

E and F are each equally distant from

A and B; and consequently, the line EF bisects AB (B. I., P. XVI., C.).

PROBLEM II.

To erect a perpendicular to a given straight line, at a given point of that line.

Let EF be a given line, and let A be a given point of that line.

From A, lay off the equal distances AB and AC; from B and C, as centres, with a radius greater than one half

E

A

F