PROP. 19, THEOR..

If a right line be a tangent to a circle, the right line perpendicular to it from the point of contact, passes through the centre of the circle.

For if possible let the centre be without this line, join it and the point of contact; this joining line is at right angles to the tangent, (prop. 18, b. 3,).. the angle contained by it and tangent is to that contained by the given line and tangent, a part to the whole, which is ab surd,.. this point is not the centre, and in the same manner it can be proved that no other point without the line at right angles to the tangent, can be the centre of that eircle.

PROP. 20, THEOR.

The angle at the centre of a circle is double the angle at the circumference, when they have the same part of the circumference for their base.

1. Let one side of the angle at the circumference pass through the centre; then because in the triangle formed by two radii and the line not passing through the centre, those radii are, the angles contained by each of them and the line not passing through the centre are, but the angle at the centre is to their sum, (prop. 32, b. 1,) .. it is double of that at the circumference.

2. Let the angle at the centre fall within that at the circumference; connect their vertices, and produce the line connecting them to the circumference, then this produced line divides the angle at the centre, and also that at the circumference; and each of those at the centre is double of its respectively internal or opposite at the circumference, (by 1st case,). &c. &c.

3. Let a side about the angle at the circumference cut a side about that at the centre, connect their vertices and produce the line connecting them to the circumference.

Then the angle (at the centre) contained by the produced part and cut side, is double of that (at the circumference) contained by the connecting line and side con

terminous with the cut side, and that at the centre contained by the produced part and uncut side, is double of that at the circumference contained by the connecting line and cutting side; if those latter angles be taken away the remainder of that at the centre, viz. the given angle at it, shall be double of the remainder of that at the circumference, viz. the given angle at it, ... &c. &c.

PROP, 21, THEOR,

The angles in the same segment of a circle are equal,

1. Let the segment be greater than a semicircle, connect the centre with the extremities of the base of those angles.

Then the angle at the centre formed by those connecting lines, is double of each of the given angles, (prop. 20, b. 3.).. the given angles are

another. (Ax. 7.)

to one

2. Let the segment be a semicircle or less than a semicircle: draw a right line from the vertex of one angle through the centre to meet the opposite circumference, and connect the vertex of the other angle with the point in which this line meets the circumference. Then those drawn lines divide the given angles into angles that are respectively

to one another, being angles in segments greater than semicircles, having the same parts of the circumference as their bases, .. the angles into which one of them is divided taken together, are to those into which the other is divided taken together, .. the given angles are -, .. &c. &c.

Cor. If two equal angles stand upon the same arch, and the vertex of one of them be in the circumference of the circle, the vertex of the other shall be in the same eircumference.

For if it be possible let the vertex of the other angle fall within or without the circumference, connect the point where a side about it or that side produced cuts the circumference, with the other extremity of the arch on which the given angles stand.

Then, because the angle thus formed, stand on the same arch with the given angle at the circumference, it is to it and.. to the other given angle (to which it is external or internal) contrary to prop. 18, b. 1.

PROP. 22, THEOR.

The opposite angles of a quadrilateral circle inscribed in a circle, are together equal to two right angles.

For assuming either triangle into which one of the diagonals resolves the quadrilateral, it has one angle, an angle of the quadrilateral, and the other two to the opposite angle of the quadrilateral, since they are separately to the parts into which the diagonal resolves it.

Cor. If one of the sides of a quadrilateral figure inscribed in a circle be produced, the external angle is equal to the internal remote angle, for each of these together with the internal adjacent angle, is equal to two right angles.

NOTE 1.-From this proposition it is evident that the angle in any segment of a circle, is to the sum of the angles contained by its subtense, and the right lines drawn from its extremities to any one point in the opposite seg

ment.

2. It is also evident that if one side of any triangle inscribed in a circle be produced, the exterual angle is to the angle in the segment at the other side of the line subtending its adjacent internal angle.

PROP. 23. THEOR.

Upon the same right line, and upon the same side of it, two similar segments of circles cannot be constructed which do not coincide.

For if it be possible, let, two similar segments be constructed, having a point in one of them without the other, connect this point with the extremities of the line on which they stand, and draw from the point in which either of those connecting lines cuts the inner seg. a right line to the opposite extremity.

Then since the segments are similar, the angles they contain are, (def. 10, b. 3,) but one of them (viz. that

in the inner seg.) is greater than the other (viz. at that the point which falls outside, (prop. 16. b. Ì,) which is absurd,.. no point in either of the given segments can fall without the other, .. &c.

PROP. 24, THEOR.

Similar segments of circles standing upon equal right lines are equal.

For if the right lines be so applied to one another that an extremity of one may fall on an extremity of the other, the other extremities must fall on one another, .`. the right lines coincide, (ax. 10,).. the segments themselves coincide and therefore are =.

PROP. 25, THEOR.

A segment of a circle being given to describe the circle of which it is the segment.

From any point in the given segment inflect two right lines, bisect them and draw through the points of bisection lines at right angles to them, produce those lines to meet; the point in which they do meet is the centre.

Because one of the inflected right lines being bisected by a perpendicular, this perpendicular passes through the centre, (prop. 1, b. 3,) likewise the other perpendicular passes through the centre, .. the centre must be the intersection of those perpendiculars.

PROP. 26, THEOR.

In equal circles, equal angles, whether they be at the centres or circumferences, stand upon equal arches.

First, let the given angles be at the centres, draw from the extremities of the arches on which they stand in each, right lines to any point in the circumference, and connect the extremities of the arches in each.

Then because the angles at the centres are, and the sides about them are respectively, the lines on which the stand are =, (prop. 4, b. 1,) i. e. the lines con

necting the extremities of the arches; also the angles at the circumferences are; (prop. 20, b. 3, & ax.7,) . ́. the segments that contain them are similar, but they stand on

right lines, .. those segments are; take away those equals from the entire circles, and the remainder, viz. the arches on which they stand are →

In the same manner it can be proved that the arches are, if the given angles at the circumferences are acute, by drawing right lines from the extremities of the arches to the centre.

But if the given angles at the circumferences be right or obtuse, bisect them, and the halves of them are, and it can be proved as above, that the arches upon which those halves stand are, whence it follows that the arches on which the given angles stand are equal.

PROP. 27. THEOR.

In equal circles, the angles which stand upon equal arches are equal, whether they be at the centres or circumferences.

For if it be possible, let one of thein be greater; cut from it a part to the other.

Then because in the given circles, the cut off angle in one is to the given angle in the other, .. the arches on which they stand are, (prop. 26, b. 3,) but the arches on which the original angles stand are also,.. the arch on which the cut off angle stands is to the arch on which the angle it is cut off from stands a part to the whole, which is absurd, .. &c.

Cor. From the preceding proposition, it is evident that in a circle, right lines which intercept equal arches are parallel, because the alternate angles are equal, and

vice versa.

PROP. 28, THEOR.

In equal circles, equal right lines cut off equal arches, the greater equal to the greater, and the less to the less.

If the equal right lines be diameters the proposition is evident.