Of the Circle

Let the chords AB and CD be parallel: then will the arcs AC and BD be equal.

D

For, draw the line AD. Then, because the lines AB and CD are parallel, the alternate angles ADC and DAB will be B

equal (Bk. I. Th. xii). But the angle

ADC is measured by half the arc AC,

and the angle DAB by half the arc BD (Th. viii): hence, the two arcs AC and BD are themselves equal.

THEOREM XIV.

The angle formed by a tangent and a chord, is measured by half the arc of the chord.

Let BAE be tangent to the circle at the

point A, and AC any chord.

From A, the point of contact, draw the diameter AD.

M

Then, the angle BAD will be a right angle (Th. v. Cor), and therefore will be measured by half the semicircle AMD B (Bk. I, Th. i. Cor. 2).

E

But the angle DAC being at the circumference, is measured by half the arc DC: hence, by the addition of equals, the two angles BAD and DAC, or the entire angle BAC will be measured by half the arc AMDC.

It may be shown, by taking the difference between the two angles DAE and DAC, that the angle CAE is measured by half the arc AC included between its sides.

5

Of the Circle.

THEOREM XV.

If a tangent and a chord are parallel to each other, they will intercept equal arcs.

Let the tangent ABC be parallel to the chord DF: then will the intercepted arcs DB, BF, be equal to each other.

For, draw the chord DB. Then, since AC and DF are parallel, the angle ABD will be equal to the angle BDF. But ABD being formed by a tangent and a chord, will be measured by half the arc

A

B

D

DB; and BDF being an angle at the circumference will be measured by half the arc BF (Th. viii). But since the angles are equal, the arcs will be equal: hence DB is equal to BF.

THEOREM XVI.

The angle formed within a circle by the intersection of two chords, is measured by half the sum of the intercepted arcs.

Let the two chords AB and CD intersect each other at the point E: then will the angle AEC, or its equal DEB, be measured by half the sum of the intercepted arcs AC, DB.

For, draw the chord AF parallel to CD. Then because of the parallels, the

B

angle DEB will be equal to the angle FAB (Bk I. Th. xiv), and the arc FD to the arc AC. But the angle FAB is measured by half the arc FDB, that is, by half the sum of the arcs FD, DB. Now, since FD is equal to AC, it follows that the angle DEB, or its equal AEC, will be measured by half the sum of the arcs DB and AC

Of the Circle.

THEOREM XVII.

The angle formed without a circle by the intersection of two secants is measured by half the difference of the intercepted

arcs.

Let the two secants DE and EB intersect each other at E: then will the angle DEB be measured by half the intercepted arcs CA and DB.

Draw the chord AF parallel to ED. D Then, because AF and ED are parallel, and EB cuts them, the angles FAB and and DEB are equal (Bk. I. Th, xiv).

E

B

But the angle FAB, at the circumference, is measured by half the arc FB (Th. viii), which is the difference of the arcs DFB and CA: hence, the equal angle E is also measured by half the difference of the intercepted arcs DFB and CA.

THEOREM XVIII.

An angle formed by two tangents is measured by hat the difference of the intercepted arcs.

Let CD and DA be two tangents to the circle at the points C and A: then will the angle CDA be measured by half the difference of the intercepted arcs CEA and CFA.

For, draw the chord AF parallel to the tangent CD. Then, because the lines CD and AF are parallel, the angle BAF

E

A

B

will be equal to the angle BDC (Bk. I. Th. xiv). But the angle BAF, formed by a tangent and a chord, is measured by

Of the Circle.

half the arc AF, that is, by half the difference of CFA and CF.

But since the tangent DC and the chord AF are parallel, the arc CF is equal to the arc CA: hence the angle BAF, or its equal BDC, which is meas-, ured by half the difference of CFA and CF, is also measured by half the difference of the intercepted arcs CFA and CA.

Cor. In like manner it may be proved that the angle E, formed by a tangent and secant, is measured by half the difference of the intercepted arcs AC and DBA.

D

B

F

THEOREM XIX.

The chord of an arc of sixty degrees is equal to the radius of the circle.

Let AEB be an arc of sixty degrees and AB its chord: then will AB be equal to the radius of the circle.

For, draw the radii CB and CA. Then, since the angle ACB is at the centre, it will be measured by the arc AEB: that is, it will be equal to sixty degrees (Bk. I. Def. 29).

E

Again, since the sum of the three angles of a triangle is equal to one hundred and eighty degrees (Bk. I. Th. xvii), it

Of the Circle.

follows that the sum of the two angles A and B will be equal to one hundred and twenty degrees. But the triangle CAB is isosceles hence, the angles at the base are equal (Bk. I. Th. vi): hence, each angle is equal to sixty degrees, and consequently, the side AB is equal to AC or CB (Bk. I. Th. vi).

PROBLEMS

RELATING TO THE FIRST AND SECOND BOOKS.

THE Problems of Geometry explain the methods of con-structing or describing the geometrical figures.

For these constructions, a straight ruler and the common compasses or dividers, are all the instruments that are absolutely necessary.

DIVIDERS OR COMPASSES.

a

The dividers consist of the two legs ba, be, which turn easily about a common joint at b. The legs of the dividers