being radii of equal circles, are equal, the point A will coincide with D, and the point B with E. Therefore the arc AB must also coincide with the arc DE, or there would be points in the one or the other unequally distant from the center, which is impossible; hence the arc A B is equal to the arc D E.

Second. If the arcs AB and D E are equal, the angles ACB and D CE will be equal.

For, if these angles are not equal, let ACB be the greater, and let ACF be taken equal to DCE. From what has been shown, we shall have the arc AF equal to the arc D E. But, by hypothesis, A B is equal to DE; hence A F must be equal to A B, the part to the whole, which is impossible; hence the angle ACB is equal to the angle DCE.

PROPOSITION VI. — THEOREM.

177. The radius which is perpendicular to a chord bisects the chord, and also the arc subtended by the chord. Let the radius C E be perpendicu

lar to the chord AB; then will CE bisect the chord at D, and the arc AB at E.

D

A

B

E

Draw the radii CA and CB. Then CA and C B, with respect to the perpendicular CE, are equal oblique lines drawn to the chord AB; therefore their extremities are at equal distances from the perpendicular (Prop. XIV. Bk. I.); hence AD and DB are equal.

Again, since the triangle ACB has the sides A C and CB equal, it is isosceles; and the line CE bisects the base A B at right angles; therefore CE bisects also the angle ACB (Prop. VII. Cor. 2, Bk. I.). Since the angles A CD, D C B are equal, the arcs A E, EB are equal

(Prop. V.); hence the radius CE, which is perpendicular to the chord AB, bisects the arc A B subtended by the chord.

178. Cor. 1. Any straight line which joins the centre of the circle and the middle of the chord, or the middle of the arc, must be perpendicular to the chord.

For the perpendicular from the centre C passes through the middle, D, of the chord, and the middle, E, of the arc subtended by the chord. Now, any two of these three points in the straight line CE are sufficient to determine its position.

179. Cor. 2. A 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, bisecting at the centre the angle which the arc subtends.

PROPOSITION VII.-THEOREM.

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

Let A, B, and C be any three points not in the same straight line; one circumference can be made to pass through them, and but one.

Join AB and BC; and bisect these straight lines by the perpendiculars D E and F E. Join DF; then, the angles BDE, BFE, being each

A

E

D

B

C

a right angle, are together equal to two right angles; therefore the angles EDF, EFD are together less than two right angles; hence DE, FE, produced, must meet in some point E (Prop. XXIII. Bk. I.).

Now, since the point E lies in the perpendicular D E, it is equally distant from the two points A and B (Prop. XV. Bk. I.); and since the same point E lies in the per

pendicular FE, it is also equally distant from the two points B and C; therefore the three distances, EA, E B, EC, are equal; hence a cireumference can be described from the center E passing through the three points A, B, C.

Again, the center, lying in the perpendicular DE bisecting the chord AB, and at the same time in the perpendicular FE bisecting the chord BC (Prop. VI. Cor. 2), must be at the point of their meeting, E. Therefore, since there can be but one center, but one circumference can be made to pass through three given points.

181. Cor. Two circumferences can intersect in only two points; for, if they have three points in common, they must have the same center, and must coincide.

182. Equal chords are equally distant from the center; and, conversely, chords which are equally distant from the centre are equal.

Let A B and D E be equal chords, and C the center of the circle; and draw CF perpendicular to A B, and CG perpendicular to DE; then these perpendiculars, which measure the distance of the chords from the center, are equal.

D

A

G

F

B

Join CA and CD. Then, in the right-angled triangle CAF, CD G, the hypothenuses CA, CD are equal; and the side A F, the half of A B, is equal to the side D G, the half of DE; therefore the triangles are equal, and CF is equal to CG (Prop. XIX. Bk. I.); hence the two equal chords A B, D E are equally distant from the center.

Conversely, if the distances CF and C G are equal, the chords A B and D E are equal.

For, in the right-angled triangles ACF, DCG, the hypothenuses CA, CD are equal; and the side CF is

equal to the side CG; therefore the triangles are equal, and AF is equal to DG; hence A B, the double of AF, is equal to D E, the double of D G (Art. 34, Ax. 6).

PROPOSITION IX. —THEOREM.

183. Of two unequal chords, the less is the farther from the center.

Of the two chords DE and AH, let AH be the greater; then will DE be the farther from the center C.

E

G

D

C

0

H

F

B

Since the chord AH is greater A than the chord D E, the arc AH is greater than the arc DE (Prop. IV.). Cut off from the arc AH a part, AB, equal DE; draw CF perpendicular to this chord, C I perpendicular to AH, and C G perpendicular to D E. CF is greater than CO (Art. 34, Ax. 8), and CO than CI (Prop. XIV. Bk. I.); therefore CF is greater than CI. But CF is equal to CG, since the chords A B, DE are equal (Prop. VIII.); therefore, C G is greater than CI; hence, of two unequal chords, the less is the farther from the center.

PROPOSITION X.-THEOREM.

A E

184. A straight line perpendicular to a radius at its termination in the circumference, is a tangent to the circle. Let the straight line BD be perpendicular to the radius CA at its Btermination A; then will it be a tangent to the circle.

Draw from the centre C to BD any other straight line, as CE. Then, since CA is perpendicular to BD, it is shorter than the oblique

D

line CE (Prop. XIV. Bk. I.); hence the point E is without the circle. The same may be shown of any other point in the line BD, except the point A; therefore BD meets the circumference at A, and, being produced, does not cut it; hence B D is a tangent (Art. 161).

PROPOSITION XI.-THEOREM.

185. If a line is a tangent to a circumference, the radius drawn to the point of contact with it is perpendicular to the tangent.

Let BD be a tangent to the circumference, at the point A; then B will the radius C A be perpendicular to BD.

For every point in BD, except A, being without the circumference (Prop. X.), any line CE drawn from the center C to BD, at any

A E

D

point other than A, must terminate at E, without the circumference; therefore the radius C A is the shortest line that can be drawn from the center to BD; hence CA is perpendicular to the tangent BD (Prop. XIV. Cor. 1, Bk. I.).

186. Cor. Only one tangent can be drawn through the same point in a circumference; for two lines cannot both be perpendicular to a radius at the same point.

PROPOSITION XII.—THEOREM.

187. Two parallel straight lines intercept equal arcs of the circumference.

First. When the two parallels are secants, as AB, DE. Draw the radius C H perpendicular to A B; and it will also be perpendicular to D E (Prop. XXII. Cor., Bk. I.);