which is therefore greater than BA; and it is opposite to the angle BAC, which is greater than the angle BCA, agreeably to the proposition.

In the same way it may be shown, that the greater angle must be opposite to the greater side.

PROP. VI, fig. 20. If any side of a triangle be produced, the exterior angle it forms with the adjacent side, is equal to the sum of the two interior and opposite angles. If the side AC of the triangle ABC be produced to D, the exterior angle BCD formed by this produced side CD, and the adjacent side of the triangle CB, will be equal to the sum of the two inward and opposite angles CAB and CBA. Let the line CE be drawn from the point C, parallel to the opposite side of the triangle AB. From what was said in Prop. 2, fig. 16, of the properties of lines intersecting each other, it appears that the vertical angles formed by such intersection. are equal to each other; and all lines intersecting one of these lines, but parallel to the other, will, with the first line, form angles respectively equal to the former; in fig. 16, the angle BEC is equal to AED; in the same way, in fig. 20, the angle ECD will be equal to ABC; and AB and EC being parallel, the angles formed by them with BC will also be equal, that is, BAC will be equal to ECD. We have now obtained the angle BCE equal to the angle at B, and ECD equal to the angle at A; but BCE and ECD together form the great exterior angle BCD, equal to the sum of the inward and opposite angles at A and B, agreeably to the enunciation of the proposition.

PROP. VII, fig. 20. The three interior angles of any triangle are equal to two right angles. The sum of the three angles at A, B, and C, of the triangle ABC, is equal to two right angles; for let the side AC be produced to D, then by the 6th Prop. the exterior angle BCD will be equal to the two inward and opposite angles at A and B. To these

equal

equal quantities let us add the remaining inward angle BCA, and we shall have the three angles of the triangle equal to the two angles BCA and BCD: but it was already shown in Prop. 1, fig. 4, that one line falling on another makes the angles on each side of the point of meeting, either two right angles, or equal to two right angles; consequently the angles BCA and BCD must be equal to two right angles: but these two angles having been shown to be equal to the sum of the three inward angles of the triangle, it follows, that the sum of these three inward angles must be equal to the sum of two right angles, as affirmed in the proposition.

1st Corollary. If two angles of a triangle, or their sum be given, the third can be found by subtracting this sum from that of two right angles.

2d Corollary. If one angle of a triangle be right, the two other angles must together be equal to one right angle.

3d Corollary. In an equilateral triangle, the angles being all equal, each is one third part of two right angles, or two thirds of one right angle.

PROP. VIII. fig. 21. The sum of all the interior angles of a polygon is equal to twice as many right angles, wanting four, as the figure has sides. Let the figure ABCDEF, be a polygon of six sides, that is, a hexagon. From any point within it, as G, draw lines to the several angles forming the six triangles AGF, AGB, &c. equal to the number of sides of the polygon. Then the sum of all the angles of each triangle (including that at the point G) will, by Prop. 7, be equal to two right angles; therefore the sum of all the angles of all the triangles will be equal to twice as many right angles as there are triangles, that is, as the figure has sides, but the sum of all the angles of all the triangles is equal to the sum of all the angles of the polygon, together with the angles at the central point G, which by 2d Corollary of Prop. 1, are shown to be equal to four right angles;

3 A

angles; therefore, subtracting these four right angles, we have all the angles of the polygon equal to twice as many right angles, excepting four, as the polygon has sides.

Having thus pointed out some of the properties of right lined figures, the next subject of consideration will be those of curve lined figures, particularly of the circle.

The characteristic of a circle is, that every part of its extent, called the circumference or pariphery, is equally distant from a point within it called the centre.

Every right line drawn from the centre to the circumference, is called a radius, as CA, CB, CH, and CD, in fig. 22; and any right line passing through the centre, and terminated both ways by the circumference, is called a diameter, as ACD, which is double the length of the radius, for AC being equal to CD, the whole AC must be double either of the parts AC or CD.

All straight lines drawn in a circle and terminated at each end by the circumference, but not passing through the centre, are termed chords; such as AG, and HD.

An arch or arc is any portion of the circumference of a circle; as the small portion AG, or the great portion ABHDG: the curve line HD is also an arch of the circle.

A figure contained between an arch of a circle and the chord or line joining the extremities of the arch, is called a segment of a circle: thus, the figure comprehended between the curve line AG and the chord AG is a segment; and, on the other hand, the figure contained within the same chord AG and the great curve ABHDG, is also a segment.

If the figure be formed by a line passing through the centre, that is, by a diameter, the segments are equal to one another, each being one half of the circle, or a semicircle, as ABD and AGD.

A figure contained within an arch of a circle, as BH, the

radius BC, and the radius HC, is called a sector. HCD, and ACH, are also sectors; but if the one radius be perpendicular to the other, that is,. if they form a right angle at the centre, as BC and DC, then the sector is called a quadrant, as being the fourth part of the whole circle, and the half of a semicircle; for AD, the diameter, being equally divided in the centre C, and CB being perpendicular to the diameter at that point, any point in CB will be equally distant from A and B; the arch AB will therefore be equal to the arch BD, and the segment ABC to the segment BDC; each of them will, consequently, be one half of the semicircle ABD, that is, each will be one fourth of the whole circle, or a quadrant.

A right line can meet the circumference of a circle only in two points, as AG or HD.

In the same circle, or in circles equal to each other, the chords of equal arches are equal to each other; and vice versa, the arches subtended by equal chords are equal to each other.

In the same or in equal circles, the greater arch is subtended by the greater chord, and the less arch is subtended by the less chord; unless the arch be greater than a semicircle, when the greater the arch the smaller the chord.

PROP. IX. fig. 23. If three points, as ABD, be taken in the circumference of a circle whose centre is C, no other circle can be drawn through the same three points not coinciding with the given circle. Join the three points by the two chords AB and BD, and from the centre C draw CE and CF perpendicular to these chords, and consequently bisecting them, or dividing them severally into two equal parts; in which case the centre of any circle passing through the points A and B, will be somewhere in the line EC; and in the same way the centre of any circle passing through the points B and D mast be somewhere in the line

FC: consequently the centre of a circle passing through the three points AB and D must be where these two lines EC and FC meet, that is, in the point C; but this is the centre of the circle in whose circumference the three given points were taken; therefore only one circle can be drawn to pass through any three given points: and hence it follows, that circles can cut one another only in two points.

PROP. X. fig. 24. The angle formed by the lines AC and BC drawn from the extremities of an arch of a circle. AB to the centre C, is double the angle formed by the lines AD and BD drawn from the same extremities to any point, as D in the opposite circumference. From D draw through the centre C the diameter DE. In the triangle DCA the sides DC and CA are equal to each other, each being a radius of the circle; the angles opposite to them are therefore equal, that is, CAD is equal to CDA: but, by the 7th Prop. it appears, that the exterior angle ACE, formed by producing the side DC, is equal to the two inward and opposite angles of the triangle, that is, since they are equal to one another, ACE must be double ADC or ADE; again, in the triangle CDB, by the same Proposition, it appears that the exterior angle ECB is equal to the two inward and opposite angles CDB and CBD; but these angles being opposite to equal sides, must be also equal; and the exterior angle ECB will be double CDB, that is, EDB : it follows, therefore, that the whole angle at the centre ACB, will be double the whole angle at the circumference ADB.

Corollary. All angles, however situated, of the same segment of a circle, or in other words, all angles formed by lines drawn from the extremities of the arch of the segment, and meeting in the circumference, are equal.

From this proposition it follows, that the angles in a semicircle are all right angles; that the angles in a segment

greater