ВО ОК III.

THE CIRCLE, AND THE MEASUREMENT

OF

ANGLES.

DEFINITIONS.

1. A CIRCLE is a plane figure, bounded by a curved line, every point of which is equally distant from a point within, called the centre.

The bounding line is called the circumference.

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2. A RADIUS is a straight line drawn from the centre to any point of the circumference.

3. A DIAMETER is a straight line drawn through the centre and terminating in the circumference.

All radii of the same circle are equal. All diameters are also equal, and each is double the radius.

4. An ARC is any part of a circumference.

5. A CHORD is a straight line joining the extremities of

an arc.

Any chord belongs to two arcs: the smaller one is meant, unless the contrary is expressed.

6. A SEGMENT is a part of a circle included between an arc and its chord..

7. A SECTOR is a part of a circle included within an an arc and the radii drawn to its extremities.

8. An INSCRIBED ANGLE is an angle whose vertex is in the circumference, and whose sides are chords.

9. An INSCRIBED POLYGON is a polygon whose vertices are in the circumference, and whose sides are chords.

10. A SECANT is a straight line which cuts the circumference in two points.

11. A TANGENT is a straight line which touches the circumference in one point. This point is called, the point of contact, or, the point of tangency.

12. Two circles are tangent to each other, when they touch each other in one point. This point is called, the point of contact, or the point of tangency.

13. A Polygon is circumscribed about a circle, when all of its sides are tangent o the circumference.

14. A Circle is inscribed in a polygon, when its circumference touches all of the

sides of the polygon.

POSTULATE.

A circumference can be described from any point as a

centre, and with any radius.

PROPOSITION I. THEOREM.

Any diameter divides the circle, and also its circumference, into two equal parts.

then will they coincide; otherwise there would be some points in either one or the other of the curves unequally distant from the centre; which is impossible (D. 1): hence, AB divides the circle, and also its circumference, into two equal parts; which was to be proved.

PROPOSITION II. THEOREM.

A diameter is greater than any other chord.

Let AD be a chord, and AB a diameter through one extremity, as A: then will AB be greater than AD.

PROPOSITION III. THEOREM.

A straight line cannot meet a circumference in more than two points.

Let AEBF be a circumference, and AB a straight line: then AB cannot meet the circumference in more than two points.

For, suppose that they could meet in three points.

We should then have three

F

A

B

E

equal straight lines drawn from the same point to the same straight line; which is impossible (B. I., P. XV., C. 2) : hence, AB cannot meet the circumference in more than two points; which was to be proved.

PROPOSITION IV. THEOREM.

In equal circles, equal arcs are subtended by equal chords; and conversely, equal chords subtend equal arcs.

If the semi-circle

Draw the diameters AB and EF ADB be applied to the semi-circle EGF, it will coincide. with it, and the semi-circumference ADB will coincide with the semi-circumference EGF But the part AMD is equal to the part ENG, by hypothesis: hence, the point D will fall on

therefore, the chord AD will coincide with

EG (A. 11), and is, therefore, equal to it; which was to be proved.

2o. Let the chords AD and EG the arcs AMD and ENG be equal.

be equal then will

The triangles ACD

Draw the radii CD and OG. and EOG have all the sides of the one equal to the corresponding sides of the other; they are, therefore, equal in all their parts: hence, the angle ACD is equal to EOG. If, now, the sector ACD be placed upon the sector EOG, so that the angle ACD shall coincide with the angle EOG, the sectors will coincide throughout; and, consequently, the arcs AMD and ENG will coincide: hence, they will be equal; which was to be proved.

PROPOSITION V. THEOREM.

In equal circles, a greater arc is subtended by a greater chord; and conversely, a greater chord subtends a greater

For, place the circle EGK upon AHL, so that the centre shall fall upon the centre C, and the point E upon A; then, because the arc EGP is greater than AMD, the point P will fall at some point H, beyond D, and the chord EP will take the position AH.

Draw the radii CA, CD, and CH. Now, the sides AC, CH, of the triangle ACH, are equal to the sides AC, CD, of the triangle ACD, and the angle ACH is