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

DAB hence, the two angles, taken together, are

mea

sured 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.

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).

A

C

E

But, the angle EDA is measured by

D

B

half the arc AC, and EAD by half the

are DB (P. XVIII.): hence, the angle DEB is measured by half the sum of the arcs AC and DB; which was to be proved.

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 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.

PROPOSITION XXI. THEOREM.

it at

An angle formed by a tangent and a chord meeting it 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 will the angle BAC be measured

by half of the arc AMC.

M

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

B

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.

6

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 will EF bisect 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.).

A

C B

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

of BC, describe arcs intersecting at D; draw the ine AD: then will AD be the perpendicular required. For, D and A are each equally distant from B and C; consequently, DA is perpendicular to BC at the given point A (B. I., P. XVI., C.).

PROBLEM III.

To draw a perpendicular to a given straight line, from a given point without that line.

and

A

Let BD be the given line, and A the given point. From A, as a centre, with a ra dius sufficiently great, describe an arc cutting BD in two points, B D; with B and D as centres, and a radius greater than one-half of RD, describe arcs intersecting at E; draw AE: then will AE be the perpendi

cular required. For, A and E are each equally distant from B and D : hence, AE is perpendicular to BD (B. I., P. XVI., C.).

PROBLEM IV.

At a point on a given straight line, to construct an angle equal to a given angle.

Let A be the given point, AB the given line, and

ing in the sides of the angle.

From A as a centre, with a radius AB, equal to KI,