364. A segment is the part of a circle comprised between an arc and its chord, as EFGE.

365. It follows from the definition of a circumference that the arc of a sector does not cease to coincide with the circumference when the sector is made to turn about its

centre.

366. THEOREM. In the same circle, or in equal circles, equal arcs are subtended by equal chords.

Let the arcs AMB and CND be equal (Fig. 17).

Turn the arc AMB about the centre O until it coincides with CND. The two arcs will then have the same extremities, and the chords AB and CD will coincide and be equal.

367. THEOREM. In the same circle, or in equal circles, equal chords are equally distant from the centre.

Let AB and CD be two equal chords in the two equal circles whose centres are O and Q. It is required to show that the perpendicular OE let fall from the centre O upon AB is equal to the perpendicular QF let fall from the centre Q upon CD.

Place the centre O upon Q. The two circles will coincide. Turn the circle whose centre is O about Q until the chord AB falls exactly upon its equal CD; then the perpendicular OE will coincide with, and be equal to, QF; for, from a point without a straight line, only one perpendicular to that line can be drawn.

368. THEOREM. The diameter perpendicular to a chord divides the chord and each of the arcs which it subtends into two equal parts.

Let AB be a chord, and MN a diameter perpendicular to the chord. which passes through the centre O and intersects the chord in D.

It is required to show

AD = DB,

M

A

B

N

FIG. 19.

Turn the semicircle NAM about MN as an axis until it falls upon the semicircle NBM.

Then, since ODA and ODB are right angles, DA will take the direction of DB, and the point A will fall somewhere in the line DB; but the point A, being in the semicircumference MAN, will fall somewhere in the semi-circumference MBN. That is, the point A, falling at the same time in the line DB and the semi-circumference MBN, must fall upon a point common to them, namely, B. So that DA will coincide with DB, arc MA with arc MB, and arc NA with arc NB.

369. A perpendicular to a chord at its middle point passes through the centre of the circle, and bisects each of the arcs which the chord subtends.

DIVISION OF THE CIRCUMFERENCE.

370. From § 178 it is seen that the whole angular magnitude about a point in a plane is divided into 360 equal parts, called degrees. Each degree is subdivided into 60 equal parts, called minutes, and each minute into 60 equal parts, called seconds.

Likewise the circumference of a circle is divided into 360

equal arcs, each corresponding to an angle of 1° at the centre. These arcs are called degrees, and are each subdivided like the angle-degree into minutes and seconds. Hence,

371. An angle is said to be measured by the arc described from its vertex as centre, and included between its sides.

The meaning of this is that the angle is such a part of 360° of angular magnitude as the corresponding arc is of 360° of circumference.

372. If the two arcs described from the vertices of two angles as centres with equal radii and included by their sides are equal, the angles are equal.

373. Having given a circumference divided into equal parts, if the several points of division are joined to the centre, the radii thus obtained will divide likewise into equal parts every circumference that has the same centre as the given circumference.

Consider for example the two concentric circumferences ABCD and abed. If the arcs AB, BC, CD are equal,

the radii OA, OB, OC will cut off equal arcs ab, bc, cd, upon the circumference abcd.

374. If we have then a circle marked off in degrees, we can easily divide into degrees a circle of any radius. This divided circle is called a protractor.

The protractor (Fig. 21) is a semicircle made of horn or hard rubber, and divided into degrees from 0° to 180°.

375. PROBLEM. Having given an angle AOB, find with the protractor the number of degrees it contains.

B

Place the diameter of the protractor upon the line OA, with its centre upon the vertex O. Read from the protractor the degrees contained between OA and OB. The number of degrees comprised between the sides OA and OB is the number o required.

FIG. 22.

A

376. The ratio of two angles is found by finding the number of degrees in each angle, and taking the ratio of the two numbers thus found.

377. PROBLEM. At a given point in a given straight line, to construct an angle equal to a given angle, by means of compasses.

Let E be the given point in the line EF, and C the given angle (Fig. 23). It is required to construct an angle at E equal to the angle C.

From Cas a centre, with any radius,

terminat

E

H

FIG. 23.

as CB, describe the arc BA, ing in the sides of the angle; and from E as a centre, with a radius equal to CB, describe the arc FH

From Fas a centre, with a radius equal to the distance BA, describe an arc intersecting the arc FH at D.

Draw DE; and the angle DEF= angle ACB.

For the chords AB and DF are equal, and, therefore, the arcs AB and DF are equal. Hence the angles E and Care equal (§ 372).

378. PROBLEM. To bisect a given arc.

A

B

FIG. 24.

It is required to bisect the arc AOB (Fig. 24).

Draw the chord AB.

From A and B as centres, with equal radii, describe arcs intersecting at E and C.

Draw EC.

EC bisects the arc AOB.

For, E and C, being two points at equal distances from A and B, determine the position of the perpendicular to the middle of

chord AB; and a perpendicular erected at the middle of a chord passes through the centre of the circle, and bisects the arc of the chord (§ 369).

PARALLEL LINES.

379. Two straight lines lying in the same plane are parallel when they cannot meet, however far produced.

B

D

To obtain two straight parallel lines, it is only necessary to draw two perpendiculars to the same straight line.

Thus, the straight lines AB and CD, being two perpendiculars drawn in a plane to the same straight line AC, are parallel; for, if they could meet, we could, from their point of intersection, let fall two distinct perpendiculars upon AC, but this is impossible; for, from a point without a straight line, only one perpendicular can be drawn from the point to the straight line.

A

FIG. 25.

C