But the angles at the centre are in proportion to the arcs on which they stand, for the arcs are their measures, and measures are the foundations of all proportion. Therefore, again,

Any angle at the centre is double the angle at the circumference, which stands on the same arc.

Refer again to the diagram. The lines A D, B D, and c D, are drawn from the angles A B and c, to n the centre of the circle; and they stand on the same arcs as the angles of the triangle. ABcand ADC both stand on the arc b; acband ADB on the arcc; and Bacand BDC on the arca. Therefore ADC is double A BC; ADB is double ACB; and BDC is double вас

The fact of the angle at the centre being double the angle at the circumference, might have been arrived at by applying the principle of symmetrical triangles to a particular case; but the general investigation is much more satisfactory.

10. It will be seen that the magnitude of the angle at the circumference increases with the arc or segment upon which it stands, and not with that in which it is contained. It is always half the angle at the centre; and as the angle at the centre is in proportion to the arc on which it stands, the angle at the circumference, which is the half of it, must always be in proportion to half that arc.

The angle in a segment is, therefore, always inversely as the segment which contains it, and directly as the segment upon which it stands. The words "segment" and "arc" being synonymous in this use of them.

These two segments, namely, the one that contains the angle, and the one upon which the angle stands, and which determines the magnitude of the angle, always between them make up the whole circumference. But the angles standing on the whole circumference, and having their vertices at the circumference, are always equal to two right angles. There

fore, if the circumference of a circle is divided into any two segments, the angles in those segments are always together equal to two right angles, as they are the supplements of each other.

Thus, if any circle, A B CD, is divided into two segments by

any chord, as for instance the chord A c, the angles ABC in the one segment, and a DCin the other, are always equal to two right angles in which way soever the chord may divide the the circle. Also, if another chord is drawn, joining the vertices of the angles, as the dotted line D B, the angles DA B, D CB, are together equal to two right angles. Hence, we have this general conclusion :

If a four-sided figure can be inscribed in a circle, that is, if a circle can be drawn so as to touch all its angles, the opposite angles of that figure are together equal to two right angles.

But this is not a property of all four-sided figures, for there is a condition in it, and we must have the means of knowing whether any given figure has or has not this condition; and this takes the form of the following problem:

To determine whether a given four-sided figure can or cannot be inscribed in a circle. As all the four sides are here placed in a circle, the line which bisects each of them at right angles must pass through the centre, and unless these lines all meet in one point, the figure cannot be inscribed in a circle. We shall take a figure at random:

The first of these is the figure made at random; the second is the trial, by bisecting the sides at right angles, as formerly explained; and as there are four crossings as marked by the dots, the figure cannot be exactly inscribed in a circle; but it can be so very nearly, as the dots are close together; and a circle drawn from the fifth dot in the centre touches the angles B and c, but it is a little within at a, and a little without at D.

Another truth which follows from the same principles is the following:

All angles in the same segment of a circle are equal to each other, in what places soever their vertices may be situated, and whether the segment be part of a larger circle or a smaller.

Thus if ABD and abd be equal segments, that is, each containing the same part of a circumference, but the one part of a larger circle than the other; then the angles which have their vertices at 1, 2 and 3 in the first, and at 4 and 5 in the

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ANGLES IN SEGMENTS.

second, and also all possible angles that could be made in either segment or in any other equal segment whatever, are exactly the same portions of a circumference. For,

Find c and c, the centres of the segments, and draw the dotted lines A C, CD in the one, and a c,c d in the other; and as the segments are the same parts of the circumference, the angles A CD, and a c d are equal, and so are the re-entering angles on the opposite sides of the dotted lines, which are their supplements to four right angles. But these supplements are double of any of the angles in the segments, for they are the angles at the centres answering to the arcs on which the supplements stand. Therefore, all the angles are halves of equal quantities; consequently the angles themselves are all equal.

Any angle in a semicircle is a right angle, because it is equal to the angle in the opposite semi-circle, and the angles in the two segments which make up an entire circumference are always together equal to two right angles.

For the same reason, an angle in a segment less than a semicircle is always greater than a right angle; and an angle in a segment greater than a semicircle is always less than a right angle. The difference of the angle from a right angle is always half the difference of the arc from a semicircle; but it is less in the case of greater, and greater in the case of less.

11. We are now prepared to solve the following problem:

Upon a given straight line, to describe a segment of a circle, which shall contain an angle equal to a given angle.

This problem is often of much service, to those who are handy with a pair of compasses, in the construction of plans; but before we proceed to it, we must premise another :

To make, at a point in a given line, an angle equal to a given angle.

This is nearly self-evident: let a be the angle, and B the point in the line. On a describe an arc cn, and with the same radius describe an arc from Bas a centre and on the side of the line, and toward that hand where the opening of the angle is to be. Then with a radius equal to the chord of c D, and from the point where the second arc meets the line, describe another arc cutting that one; draw a line through the point B and the intersection of the arcs, and the angle is made.

Draw the chords CD and E F, and they are equal, being chords of equal arcs; and the other sides of the triangles are all made equal. Therefore, the angle at B is equal to the angle

at A.

Let us now return to our main problem: given any line

it is required to construct upon it a segment of a circle that shall contain an angle equal to a sixth part of a circumference, or 60°. This is the angle of an equilateral triangle, for it is the third part of two right angles. The segment which we want must be greater than a semicircle, because the angle in it is less than a right angle. The given line A B is the chord of the arc on which the segment has to stand; and therefore it must subtend at the centre 120°, which is double the angle in the segment.

The problem is thus reduced to this: To apply to the given line A B a triangle which shall have its other two sides equal to each other, and the angle between them equal to 120°, or one third of a circumference; and the point of this angle will