10. Two circumferences touch each other when they meet, but do not cut one another.

11. A polygon is described about a circle, when each side of the polygon touches the circumference of the circle.

In the same case, the circle is said to be inscribed in the polygon.

PROPOSITION I. THEOREM.

Every diameter divides the circle and its circumference into two equal parts.

Let ACBD be a circle, and AB its diameter.

A

B

The line AB divides the circle and its circumference into two equal parts. For, if the figure ADB be applied to the figure ACB, while the line AB remains common to both, the curve line ACB must coincide exactly with the curve line ADB. For, if any part of the curve ACB were to fall either within or without the curve ADB, there would be points in one or the other unequally distant from the center which is contrary to the definition of a circle. Therefore, every diameter, &c.

D

PROPOSITION II. THEOREM.

A straight line can not meet the circumference of a circle tn more than two points.

For, if it is possible, let the straight line ADB meet the circumference CDE in three points, C, D, E. Take F, the center of the circle, and join FC, FD, FE. Then, because F is the center of the circle, the three straight lines FC, FD, FE are all equal to each other; hence, three equal straight lines have

F

B

D

been drawn from the same point to the same straight line,

which is impossible (Prop. XVII., Cor. 2, Book I.). Therefore, a straight line, &c.

PROPOSITION III. THEOREM.

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

being equal to the diameter EF, the semicircle ADB may be applied exactly to the semicircle EHF, and the curve line AIDB will coincide entirely with the curve line EMHF (Prop. I.). But the arc AID is, by hypothesis, equal to the arc EMH; hence the point D will fall on the point H, and therefore the chord AD is equal to the chord EH (Axiom 11, B. I.).

Conversely, if the chord AD is equal to the chord EH, then the arc AID will be equal to the arc EMH.

For, if the radii CD, GH are drawn, the two triangles ACD, EGH will have their three sides equal, each to each viz.: AC to EG, CD to GH, and AD equal to EH; the tri angles are consequently equal (Prop. XV., B. I.), and the an gle ACD is equal to the angle EGH. Let, now, the semicircle ADB be applied to the semicircle EHF, so that AC may coincide with EG; then, since the angle ACD is equal to the angle EGH, the radius CD will coincide with the radius GH, and the point D with the point H. Therefore, the arc AID must coincide with the arc EMH, and be equal to it. Hence, in equal circles, &c.

PROPOSITION IV. THEOREM.

In equal circles, equal angles at the center, are subtended by equal arcs; and, conversely, equal arcs subtend equal angles at the center.

Let AGB, DHE be two equal circles, and let ACB, DFE be equal angles at their centers; then will the arc AB be equal to the arc DE. Join AB, DE; and, because the cir

is equal to the chord DE, the arc AB must be equal to the arc DE (Prop. III.).

Conversely, if the arc AB is equal to the arc DE, the angle ACB will be equal to the angle DFE. For, if these angles are not equal, one of them is the greater. Let ACB be the greater, and take ACI equal to DFE; then, because equal angles at the center are subtended by equal arcs, the arc AI is equal to the arc DE. But the arc AB is equal to the arc DE; therefore, the arc AI is equal to the arc AB, the less to the greater, which is impossible. Hence the angle ACB is not unequal to the angle DFE, that is, it is equal to it. Therefore, in equal circles, &c.

PROPOSITION V. THEOREM.

In the same circle, or in equal circles, a greater arc is subtended by a greater chord; and, conversely, the greater chord subtends the greater arc.

In the circle AEB, let the arc AE be greater than the arc AD; then will the chord AE be greater than the chord AD.

Draw the radii CA, CD, CE. Now, if the arc AE were equal to the arc AD, A the angle ACE would be equal to the angle ACD (Prop. IV.); hence it is clear that if the arc AE be greater than the arc

E

C

B

AD, the angle ACE must be greater than the angle ACD. But the two sides AC, CE of the triangle ACE are equal to the two AC, CD of the triangle ACD, and the angle ACE is greater than the angle ACD; therefore, the third side AE is greater than the third side AD (Prop. XIII., B. I.); hence the chord which subtends the greater arc is the greater.

Conversely, if the chord AE is greater than the chord AD, the arc AE is greater than the arc AD. For, because the two triangles ACE, ACD have two sides of the one equal to two sides of the other, each to each, but the base AE of the one is greater than the base AD of the other, therefore

the angle ACE is greater than the angle ACD (Prop. XIV. B. I.); and hence the arc AE is greater than the arc AD (Prop. IV.). Therefore, in the same circle, &c.

Scholium. The arcs here treated of are supposed to be less than a semicircumference. If they were greater, the opposite property would hold true, that is, the greater the arc the smaller the chord.

PROPOSITION VI. THEOREM.

The radius which is perpendicular to a chord, bisects the chord, and also the arc which it subtends.

Let ABG be a circle, of which AB is a chord, and CE a radius perpendicular to it; the chord AB will be bisected in D, and the arc AEB will be bisected in E.

Draw the radii CA, CB. The two rightangled triangles CDA, CDB have the side AC equal to CB, and CD common; there- A fore the triangles are equal, and the base AD is equal to the base DB (Prop. XIX., B. I.).

Secondly, since ACB is an isosceles triangle, and the line CD bisects the base at right angles, it bisects also the vertical angle ACB (Prop. X., Cor. 1, B. I.). And, since the angle ACE is equal to the angle BCE, the arc AE must be equal to the arc BE (Prop. IV.); hence the radius CE, perpendicular to the chord AB, divides the arc subtended by this chord, into two equal parts in the point E. Therefore, the radius, &c.

Scholium. The center C, the middle point D of the chord AB, and the middle point E of the arc subtended by this chord, are three points situated in a straight line perpendicular to the chord. Now two points are sufficient to determine the position of a straight line; therefore any straight ne which passes through two of these points, will necessariy pass through the third, and be perpendicular to the chord. Also, the perpendicular at the middle of a chord passes through the center of the circle, and through the middle of the arc subtended by the chord.

PROPOSITION VII. THEOREM.

Through three given points, not in the same straight line, one circumference may be made to pass, and but one.

B

Let A, B, C be three points not in the same straight line; they all lie in the circumference of the same circle. Join AB, AC, and bisect these lines by the perpendiculars DF, EF; DF and EF produced wil meet one another. For, join DE; then, because the angles ADF, AEF are together equal to two right angles, the angles FDE and FED are together less than two right angles; therefore DF and EF will meet if produced (Prop. XXIII., Cor. 2, B. I.). Let them meet in F. Since this point lies in the perpendicular DF, it is equally distant from the two points A and B (Prop. XVIII., B. I.); and, since it lies in the perpendicular EF, it is equally distant from the two points A and C; therefore the three distances FA, FB, FC are all equal; hence the circumference described from the center F with the radius FA will pass through the three given points A, B, C.

A

Secondly. No other circumference can pass through the same points. For, if there were a second, its center could not be out of the line DF, for then it would be unequally distant from A and B (Prop. XVIII., B. I.); neither could it be out of the line FE, for the same reason; therefore, it must be on both the lines DF, FE. But two straight lines can not cut each other in more than one point; hence only one circumference can pass through three given points. Therefore, through three given points, &c.

Cor. Two circumferences can not cut each other in more than two points, for, if they had three common points, they would have the same center, and would coincide with each other

PROPOSITION VIII. THEOREM.

Equal chords are equally distant from the center; and of two unequal chords, the less is the more remote from the center.

Let the chords AB, DE, in the circle ABED, be equal to ɔne another; they are equally distant from the center. Take

C