34

EQUAL ANGLES AT THE CENTRE. [BOOK II.

PROP. III. THEOREM.

In the same, or in equal circles, equal angles at the centre are subtended by equal arcs.

Let C be the centre of the circle, and let the angle ACB be equal to ECD; then we have to prove that the arcs AB, DE, which subtend these angles, are equal.

Fig. 31.

B

D

G

Draw the chords AB, DE; then since radii are all equal (Def. 4), the triangles ACB, DCE, have two sides and the included angle of the one, equal respectively to two sides and the included angle of the other, hence they are equal; so that if ACB be applied to DCE, there will be an exact coincidence (B. I. Def. 23), the point A would fall on D, and the point B on E; the two extremities therefore of the arc AB thus coinciding with those of DE, all the intermediate parts must coincide, inasmuch as they are equally distant from the centre (Def. 2).

Cor. 1. It follows moreover that equal angles at the centre are subtended by equal chords.

Cor. 2. If the angle at the centre be bisected, both the arc and chord subtending it are also bisected.

Cor. 3. If the angles are unequal the chords will be unequal.

Scholium. The above reasoning applies obviously to the case of equal circles, as one would entirely coincide with the other.

In the same circle, or in equal circles, equal arcs subtend equal angles at the centre.

Let (as in Prop. 3) the arc AB be equal to the arc DE; then we have to prove that the angles ACB, DCE, are equal.

If ACB were applied to DCE, the arcs AB, DE, being equal, would coincide, the point A would fall upon D, and B upon E, the line AB would coincide with DE; hence the two triangles ACB, DCE, having the three sides of the one respectively equal to the three sides of the other, are equal (B. I. Prop. 22), and the angles ACB, DCE, are equal.

Cor. 1. Equal chords subtend equal angles at the

centre.

Cor. 2. Therefore equal chords subtend equal arcs, and conversely equal arcs are subtended by equal chords. Cor. 3. If the chords are unequal the angles and arcs which they subtend are unequal.

PROP. V. THEOREM.

An angle at the centre, subtended by half a semi-circumference, is a right angle.

Let the point D be the middle of the semi-circumference ADB; then we have to

Fig. 32.

equal, being each half of a semi-circumference; there

36

CIRCLE THROUGH THREE POINTS. [BOOK II.

fore the angles ACD, DCB, are equal (Prop. 4); but these angles are adjacent; hence each is a right angle. (B. I. Def. 6.)

PROP. VI. THEOREM.

A line drawn from the centre of a circle perpendicular to a chord, bisects it, and also the arc which it subtends.

Let the line CE be drawn from C, the centre, perpendicular to AB; then we have to prove that it will bisect AB, and also the arc AEB.

Fig. 33.

D

B

E

Draw the radii AC, BC; these are equal (Def. 4), therefore AD=DB (B. I. Prop. 19, Cor. 2); hence the two triangles ACD, BCD, having the three sides of the one equal to the three sides of the other, are equal, (B. I. Prop. 22), and the angles ACD, BCD, are equal; and therefore the arcs AE, EB, are equal (Prop. 3, Cor. 2).

Cor. 1. Hence the line joining the centre of the circle and the middle of the chord, or the middle of the arc, is perpendicular to the chord.

Cor. 2. A perpendicular through the middle of the chord passes through the centre, and through the middle of the arc, and bisects the angle at the centre which the chord subtends.

PROP. VII. THEOREM.

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

To prove this, let A, B, D, be the given points, and

7

BOOK II.] DISTANCE OF CHORDS FROM CENTRE.

B

Fig. 34.

37

join them by the lines AB, AD; from E, F, the middle points of these lines, erect the perpendiculars EC, FC, which will meet in some point,C; for if they do not meet, they must be parallel, and then the lines AB, AD,

which are perpendicular to them, would also be parallel

A

(B. I. Prop. 10, Cor. 3), or else form but one straight line ; but they meet at A, and are therefore not parallel; and by the hypothesis BAD is not a straight line hence EC, FC, will meet; and if from the point C, in which they meet, as a centre, we describe a circumference, with the radius CA, it will pass through the three points A, B, D. For draw CA, CB, CD, and because AC, BC, meet AB at equal distances from the foot of the perpendicular CE, they are equal (B. I. Prop. 18); for the same reason AC, DC, are equal; hence the three points A, B, D, lying at equal distances from C, are in the circumference of a circle whose centre is C, and radius CA.

Again, only one circumference can pass through the three points A, B, D, because the lines EC, FC, intersect each other but once, and therefore there can be but one centre, and with the same centre and the same radius, it is obvious only one circumference can be described.

Cor. One circumference cannot cut another in more than two points.

tre.

PROP. VIII. THEOREM.

Two equal chords are equally distant from the cen

Of two unequal chords the shorter is at the greater dis

The lines OH, OK, being drawn from the centre perpendicular to the chords AB, CD, bisect them (Prop. 6); therefore AH, the half of AB, is equal to CK, the half of CD (B. I. Ax. 3). Now in the right-angled triangles AOH, CKO, AO is equal to OC, and AH to CK; hence these triangles are equal (B. I. Prop. 20), and OH OK.

2d. Let AE be longer than AB; we have to prove that AB is further from the centre than AE. Draw OI perpendicular to AE. OL is longer than OI (B. I. Prop 15, Cor.), but OL<OH; hence OI<OH; but OI, OH, measure the distances from O to AE, AB.

PROP. IX. THEOREM.

Every straight line perpendicular to a radius at its extremity is a tangent to the circle.

Conversely. Every tangent to a circle is perpendicular to the radius drawn to the point of contact.

1st. Let AB be perpendicular to the radius CD, at its extremity D; then we have to prove that AB is a tangent to the circle. Το do this we have only to show that AB cannot touch the

Fig. 36.

D E