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

C

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

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 the arcs DB and AC

um

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

gres.

Lst the two secants DE and EB intersec. 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 he difference of the intercepted arcs CEA and CFA.

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

E

D

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 measured by half the difference of CFA and CF, is also measured by half the difference of the intercepted arcs CFA and CA.

D

E

Ccr. 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.

B

THEOREM XIX.

"he 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 constructing or describing the geometrical figures.

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

DIVIDERS OR COMPASSES.

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

Problems.

are extended or brought together by placing the forefinger or the joint at b, and pressing the thumb and fingers against the legs.

PROBLEM 1.

On any line, as CD, to lay off a distance equal to AB.

Take up the dividers with the thumb and second finger, and place the forefinger on the joint at b.

Α

B

Then, raise the dividers, place one foot at C, and mark with the other the distance CE: and this distance wil! evidently be equal to AB.

PROBLEM II.

To describe from a given centre the circumference of a circle having a given radius.

Let C be the given centre, and CB the given radius.

Place one foot of the dividers at C, and extend the otner eg until it teaches to B. Then, turn the dividers around the leg at C, and the other leg will describe the required circumference.