THE CIRCLE.

SECTION II.-Angles in a Circle.

Reference to Theorem IV.-In order to obtain a numerical measure of angles, we have already introduced two propositions respecting angles in a circle-namely, the theorems that "equal angles at the centre stand on equal arcs," and that "angles at the centre, standing on equal arcs, are equal."

These propositions have not been employed in any subsequent demonstrations, and they were proved quite independently of any other propositions. As at this point, however, they fall into the general chain of reasoning, they should be here repeated.

Angles whose Vertices are at the Circumference. We have next to consider the properties of angles in a circle having their vertices at the circumference. Such angles are called angles at the circumference, and are said to be inscribed in the circle.

Take two points, A, C, on the circumference, and draw from these points lines A B, C B to the same point, B, on the circumference, and also lines A O CO to the centre O. Wherever the point B may be on the circumference ABC = half of AOC. (See Fig., Theorem XXXI.)

Now, there can be but one LAOC at the centre on the arc AC, but there may be many angles at the circumference, and each of the latter is half ▲ AO C. Consequently all angles at the circumference, on the same or on equal arcs, are equal.

As the part of a circle cut off by a chord is termed a segment, this very important property of the circle may be stated thus-"Angles in the same segment of a circle or in equal segments of equal circles are equal.

M

Again, take two fixed drawing-pins, A and B, and having cut out of paper an angle, M, place it against the points A and B; then move the paper into different positions, always keeping it in contact with the drawing-pins A and B. The vertex, M, of the angle will trace out the arc of a circle from A to B. The locus of the vertex M is the arc A M B (Fig. 95). If we take the supplement of M and use it on the other side of A B, the vertex will complete the whole circle. If the angle M be a right angle, the locus will be a semi-circumference. Not only will lines joining any point,

FIG. 95.

B

M, in the semi-circumference with A and B contain a right angle, but the lines joining with A and B a point, N, not in the circumference, will not contain a right angle. This we have seen

is implied in saying that the semi-circumference is the locus of M.

If the angle M be acute, the locus is an arc greater than a semi-circumference; and if M be obtuse, as in Fig. 95, the locus will be an arc less than a semi-circumference.

FIG. 96.

B

To Erect a Perpendicular at the end of a Line. In further illustration of the fact that the locus of the vertices of right-angled triangles on the same base is a semi-circumference, let us find a method of erecting a perpendicular at the extremity, A, of a straight lire, A B.

FIG. 97

FIG. 98.

A

First, let us take any point, O, on the paper, and then let us place one point of a pair of compasses at a point, O, so that, in describing the circle, the other may cut the straight line AB in the point A and some other point B; join B O, and produce it till it meets the circle in C, the line CA will be the perpendicular required; for, since B C is a diameter, the angle B A C inscribed in a semicircle is a right angle.

Or we may first take a point a in A B (Fig. 98), and find a point b at the same distance from A as from a. Then, if we draw the line ab and continue it to C, so that bC : = ba, and finally join C A, CA will be perpendicular to A B.

Since is at the same distance from a, A, and C, ▲ CA a is the angle in a semicircle, and is therefore a right angle.

Set-square for Drawing Angles. - Another application of the theorems of this chapter is afforded by the construction of the setsquare represented in Fig. 99, and used to make angles of 90°, 75°, 60°, 45°, 30°, 15°-angles which are frequently required.

First let us construct a circumference of from 8 to 12 inches in diameter, and mark off the radius from B to A; let us find also the

Iniddle point F of the arc BFC; join CF, and produce it tiil it meets

So that the square contains all multiples of 15° to 90°.

Four-sided Figures in a Circle.-From the property of inscribed angles it will be proved that if any four-sided figure be drawn with its four vertices on the circumference of a circle each pair of opposite angles in the figure must be supplementary.

A Circle through Four given Points.-Through three given points, which are not in the same straight line, a circle can always be drawn; but it is not always possible to describe a circle passing through four given points. It may, however, happen, with a particular set of four points, that a circle made to pass through three of them will also pass through the fourth. The theorems of this chapter furnish two tests for proving whether any four given points are so situated that a circle may be described through all of them.

Let A, B, C, D be the points, and let them be joined by four lines. First, let two of these lines, A C, B D, intersect; then, if LABD

=

LACD, the four points A, B, C, D lie on the same circumference.

Secondly, let no two of the lines intersect; then, if BCD is the supplement of 4 BAD, the points lie on the same circumference.

THEOREMS ON CHAPTER IX.

SECTION II.

DEFINITIONS.

42. Angles are said to be at the centre or circumference of a circle when their vertices are at the centre or on the circumference; and they are said to stand on the arc which is intercepted by their sides.

43. A chord divides a circle into two parts, called segments; the part containing the centre being called. the major segment, and the other the minor segment.

44. An angle in a segment is the vertical angle of a triangle whose base is the chord and whose vertex lies on the arc of the segment.

45. A sector of a circle is the figure contained by two radii and the arc they intercept.

THEOREM XXX.

The angle at the centre of a circle is double the angle at the circumference on the same arc.

There are three cases to consider, according as the centre of the circle is (1) on a side of the angle, (2) within the angle, or (3) without the angle.

First. Let ABC be the angle at the circumference on the arc A C, and let A O C be the angle at the centre on the same arc. Let the side BC pass through the centre O. Join AO. Then, because AO = BO, the angle O A BOBA. But the exterior angle A O C of the triangle AO B is equal to the two interior and opposite

angles OAR O B A, and therefore is double of one of them, A B C.

Second. Let the centre O be within the angie A B C. Draw the diameter BO D, and join A O, CO. Then, by the first case,

LAOD 24 ABD

4 COD=24CBD.

Therefore, by adding these equals, it follows that
LAOC=24 ABC.

0

Б

A D

Third.—Let O fall without ▲ ABC.

same construction as before

LAOD=24ABD

4COD=24 C B D.

And by subtracting, we have—

▲ AOC=24 ABC.

A

B

Making the

THEOREM XXXI.

Angles in the same segment are equal to one another. Let A B C, A D C be angles at the circumference on the arc A C; then shall

B

D

4 ABC = 4 ADC.

Let A and C be joined to the centre O. Then each of the angles A B C, ADC is half the angle A OC; consequently, 4 ABC= < ADC.