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THEOREM.

114. If the distance of two centres is less than the sum of the radii, and if at the same time the greater radius is less than the sum of the smaller and the distance of the centres, the two circles will cut each other.

58.

Demonstration. In order that the intersection may take place, the triangle ACD (fig. 57, 58) must be possible. It is necessary Fig. 57, then, not only that CD (fig. 57) should be less than AC+AD, Fig. 57. but also that the greater radius AD (fig. 58) should be less than Fig. 58. AC+ CD. Now, while the triangle CAD can be constructed, it is clear that the circumferences described from the centres C and D will cut each other in A and B.

THEOREM.

115. If the distance CD (fig. 59) of the centres of two circles is Fig. 59 equal to the sum of their radii CA, CD, these two circles will touch each other externally.

Demonstration. It is evident that they will have the point A common, but they can have no other, for in order that there may be two points common, it is necessary that the distance of the centres should be less than the sum of the radii (114).

THEOREM.

116. If the distance CD of the centres of two circles is equal to the difference of their radii CA, AD (fig. 60), these two circles will Fig. 60. touch each other internally.

Demonstration. In the first place it is evident, that they will have the point A common; and they can have no other, for in order that they may have two points common, it is necessary that the greater radius AD should be less than the sum of the radius AC and the distance of the centres CD (114), which is contrary to the supposition.

117. Corollary. Hence, if two circles touch each other, either internally or externally, the centres and the point of contact are in the same straight line.

118. Scholium. All the circles, which have their centres in the straight line CD and whose circumferences pass through the point A, touch each other, and have only the point A common. And if through the point A we draw AE perpendicular to CD, the straight line AE will be a tangent common to all these circles.

THEOREM.

119. In the same circle, or in equal circles, equal angles ACB, Fig. 61. DCE (fig. 61), the vertices of which are at the centre, intercept upon the circumference equal arcs AB, DE.

Reciprocally, if the arcs AB, DE, are equal, the angles ACB, DCE, also will be equal.

Demonstration. 1. If the angle ACB is equal to the angle DCE, these two angles may be placed the one upon the other, and as their sides are equal, it is evident, that the point A will fall upon D, and the point B upon E. But in this case the arc AB must also fall upon the arc DE; for if the two arcs were not coincident, there would be points in the one or the other at unequal distances from the centre, which is impossible; therefore the arc AB DE.

=

2. If we suppose AB = DE, the angle ACB will be equal to DCE; for, if these angles are not equal, let ACB be the greater, and let ACI be taken equal to DCE; and we have, according to what has just been demonstrated, AI = DE. But, by hypothesis, the arc ABDE; we should consequently have AI AB, or the part equal to the whole, which is impossible; therefore the angle ACB = DCE.

THEOREM.

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120. In the same circle, or in equal circles, if two angles at the Fig. 62. centre ACB, DCE (fig. 62), are to each other, as two entire numbers, the intercepted arcs AB, DE, will be to each other, as the same numbers, and we shall have this proportion;

angle ACB angle DCE:: arc AB: arc DE.

:

Demonstration. Let us suppose, for example, that the angles ACB, DCE, are to each other, as 7 to 4; or, which amounts to the same, that the angle M, which will serve as a common measure, is contained seven times in the angle ACB, and four times in the angle DCE. The partial angles AC m, m Cn, n Cp, &c., DC x, x Cy, &c., being equal to each other, the partial arcs Am, mn, np, &c., Dx, x y, &c., will also be equal to each other (119), and the entire arc AB will be to the entire arc DE, as 7 to 4. Now it is evident, that the same reasoning might be used, whatever numbers were substituted in the place of 7 and 4; therefore, if the ratio of the angles ACB, DCE, can be expressed by

entire numbers, the arcs AB, DE, will be to each other, as the angles ACB, DCE.

121. Scholium. Reciprocally, if the arcs AB, DE, are to each other, as two entire numbers, the angles ACB, DCE, will be to each other, as the same numbers, and we shall have always ACB: DCE :: AB: DE; for the partial arcs Am, mn, &c., Dx, xy, &c., being equal, the partial angles AC m, m Cn, &c., DC x, x Cy, &c., are also equal.

THEOREM.

122. Whatever may be the ratio of two angles ACB, ACD, (fig. 63), these two angles will always be to each other, as the arcs Fig. 68. AB, AD, intercepted between their sides and described from their vertices, as centres, with equal radii.

Demonstration. Let us suppose the less angle placed in the greater; if the proposition enunciated be not true, the angle ACB will be to the angle ACD, as the arc AB is to an arc greater or less than AD. Let this arc be supposed to be greater, and let it be represented by AO; we shall have,

angle ACB angle ACD:: arc AB : arc AO.

Let us now imagine the arc AB to be divided into equal parts, of which each shall be less than DO, there will be at least one point of division between D and O; let I be this point, and join CI; the arcs AB, AI, will be to each other, as two entire numbers, and we shall have, by the preceding theorem,

angle ACB angle ACI :: arc AB arc AI.

Comparing these two proportions together, and observing, that the antecedents are the same, we conclude that the consequents are proportional (111), namely,

angle ACD: angle ACI :: arc AO: arc AI.

But the arc AO is greater than the arc AI; it is necessary then, in order that this proportion may take place, that the angle ACD should be greater than the angle ACI; but it is less; it is therefore impossible, that the angle ACB should be to the angle ACD, as the arc AB is to an arc greater than AD.

By a process of reasoning altogether similar, it may be shown, that the fourth term of the proportion cannot be less than AD;

The reference by Roman numerals is to the Introduction.

therefore it is exactly AD; and we have the proportion

angle ACB angle ACD:: arc AB : arc AD.

123. Corollary. Since the angle at the centre of a circle and the arc intercepted between its sides have such a connexion, that when one increases or diminishes in any ratio whatever, the other increases or diminishes in the same ratio, we are authorized to establish one of these magnitudes as the measure of the other; thus we shall, in future, take the arc AB as the measure of the angle ACB. The only thing to be observed in the comparison of angles with each other is, that the arcs, which are used to measure them, must be described with equal radii. This is to be understood in the preceding propositions.

124. Scholium. It may seem more natural to measure a quantity by another quantity of the same kind, and upon this principle it would be convenient to refer all angles to the right angle; and thus, the right angle being the unit of measure, the acute angle would be expressed, by a number comprehended between O and 1, and an obtuse angle by a number between 1 and 2. But this manner of expressing angles would not be the most convenient in practice. It has been found much more simple to measure them by arcs of a circle on account of the facility of making arcs equal to given arcs and for many other reasons. Besides, if the measure of angles by the arcs of a circle be in some degree indirect, it is not the less easy to obtain, by means of them, the direct and absolute measure; for, if we compare the arc, which is used as the measure of an angle, with the fourth part of the circumference, we have the ratio of the given angle to a right angle, which is the absolute measure.

125. Scholium 11. All that has been demonstrated in the three preceding propositions, for the comparison of angles with arcs, is equally applicable to the purpose of comparing sectors with arcs; for sectors are equal, when their arcs are equal, and in general they are proportional to the angles; hence two sectors ACB, ACD, taken in the same circle or in equal circles, are to each other, as the arcs AB, AD, the bases of these sectors.

It will be perceived therefore, that the arcs of a circle, which are used as a measure of angles, will also serve as the measure of different sectors of the same circle or of equal circles.

THEOREM.

65.

126. The inscribed angle BAD (fig. 64, 65), has for its measure Fig. 64, the half of the arc BD comprehended between its sides.

Demonstration. Let us suppose, in the first place, that the centre of the circle is situated in the angle BAD (fig. 64); we Fig. 64. draw the diameter AE, and the radii CB, CD. The angle BCE, being the exterior angle of the triangle ABC, is equal to the sum of the two opposite interior angles, CAB, ABC. But, the triangle BAC being isosceles, the angle CAB = ABC; hence the angle BCE is double of BAC. The angle BCE, having its vertex at the centre, has for its measure the arc BE; therefore the angle BAC has for its measure the half of BE. For a simi lar reason the angle CAD has for its measure the half of ED; therefore BAC + CAD, or BAD, has for its measure the half BE+ED, or the half of BD.

Let us suppose, in the second place, that the centre C (fig. 65), Fig. 65. is situated without the angle BAD; then, the diameter AE being drawn, the angle BAE will have for its measure the half of BE, and the angle DAE the half of DE; hence their difference BAD will have for its measure the half of BE minus the half of ED, or the half of BD.

Therefore every inscribed angle has for its measure the half of the arc comprehended between its sides.

127. Corollary 1. All the angles BAC, BDC (fig. 66), &c., Fig. 66. inscribed in the same segment, are equal; for they have each for their measure the half of the same arc BOC.

128. Corollary 1. Every angle BAD (fig. 67), inscribed in Fig. 67. a semicircle, is a right angle; for it has for its measure the half of the semicircumference BOD, or the fourth of the circumfer

ence.

To demonstrate the same thing in another way, draw the radius AC; the triangle BAC is isosceles, and the angle

BAC-ABC;

= ADC;

the triangle CAD is also isosceles, and the angle CAD =
hence BAC + CAD, or BAD = ABD+ADB. But, if the two
angles B and D of the triangle ABD are together equal to the
third BAD, the three angles of the triangle will be equal to twice
the angle BAD; they are also equal to two right angles; there-
fore the angle BAD is a right angle.

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