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straight line, but the circumference of only one circle may be made to pass through the same points.
Demonstration. Join AB, BC, and bisect these two straight lines by the perpendiculars DE, FG ; these perpendiculars will meet in a point 0.
For the lines DE, FG, will necessarily cut each other, if they are not parallel. Let us suppose that they are parallel; the line AB perpendicular to DE will be perpendicular to FG (65), and the angle K will be a right angle; but BK, which is BD producéd, is different from BF, since the three points A, B, C, are not in the same straight line ; there are then two perpendiculars BF, BK, let fall from the same point upon the same straight line, which is impossiblc (50); therefore the perpendiculars DE, FG, will always cut each other in some point 0.
Now the point O, considered with reference to the perpendicular DE, is at equal distances from the two points A and B (55); also this same point o, considered with reference to the perpendicular FG, is at equal distances from the two points B and C; hence the three distances 0.4, OB, OC, are equal; therefore the circumference, described from the centre O with the radius OB, will pass through the three points A, B, C.
It is thus proved, that the circumference of a circle may be made to pass through any three given points, which are not in the same straight line; it remains to show, that there is only one circle, which can be so described.
If there were another circle, the circumference of which passed through the three given points A, B, C, its centre could not be without the line DE (55), since, in this case, it would be at unequal distances from A and B ; neither can it be without the line FG, for a similar reason; it will then be in both of these lines at the same time. But two lines can cut each other in only one point (32); there is therefore only one circle, whose circumference can pass through three given points.
108. Corollary. Two circumferences can meet cach other only in two points ; for, if they had three points common, they would have the same centre, and would make one and the same circumference.
109. Two equal chords are at the same distance from the centre, and of two unequal chords the less is at the greater distance from the centre.
Demonstration 1. Let the chord AB = DE (fig. 53). Bisect Fig.53. these chords by the perpendicular CF, CG, and draw the radii CA, CD.
The right-angled triangles CAF, DCG, have the hypothenuses CA, CD, equal ; moreover the side AF, the half of AB, is equal to the side DG, the half of DE; the triangles then are equal (59), and consequently the third side CF is equal to the third side CG; therefore the two equal chords AB, DE, are at the same distance from the centre.
2. Let the chord AH be greater than DE, the arc AKH will be greater than the arc DME (103). Upon the arc AKH take the part ANB = DME, draw the chord AB, and let fall the perpendicular CF upon this chord, and the perpendicular CI upon AH; CF is evidently greater than Co, and Co than CI (52); for a still stronger reason CF> CI. But CF - Ci, since the chords AB, DE, are equal. Therefore CG> CI, and of two unequal chords the less is at the greater distance from the centre,
110. The perpendicular BD (fig. 54), at the extremity of the Fig. 54: radius AC, is a tangent to the circumference.
Demonstration. Since every oblique line CE is greater than the perpendicular CA (52), the point E is without the circle, and the line BD has only the point A in common with the circumference ; therefore BD is a tangent (97).
111. Scholium. We can draw through a given point A only one tangent AD to the circumference; for, if we could Jraw another, it would not be a perpendicular to the radius CA, and with respect to this new tangent the radius CA would be an oblique line, and the perpendicular let fall from the centre upon this tangent would be less than CA; therefore this supposed tangent would pass into the circle and become a secant.
112. Two parallels AB, DE (fig. 55), intercept upon the circumference equal arcs MN, PR.
Demonstration. The proposition admits of three cases.
1. If the two parallels are secants, draw the radius CH perpendicular to the chord MP, it will also be perpendicular to its parallel NQ (64), and the point H will be at the same time the middle of the arc MHP and of NHQ (105); whence the arc MH= HP, and the arc NH = HQ; also
MH - NH = HP — HQ, that is. MN = PQ. Fig. 56.
2. If of the two parallels AB, DE (fig. 56), one be secant and the other a tangent; to the point of contact H draw the ra. dius CH; this radius will be perpendicular to the tangent DE (110), and also to its parallel MP. But, since CH is perpendicular to the chord MP, the point H is the middle of the arc MHP; therefore the arcs MH, HP, comprehended between the parallels AB, DE, are equal.
3. If the two parallels, DE, IL, are tangents, the one at H and the other at K; draw the parallel secant AB, and we shall have, according to what has just been demonstrated MH= HP, and MK = KP; therefore the entire arc HMK = HPK, and it is moreover evident, that each of these arcs is a semicircumfere
113. If the circumferences of two circles cuť each other in two points, the line which passes through their centres will be perpendicular to the chord, which joins the points of intersection, and will
bisect it. Fig. 57, Demonstration. The line AB (fig. 57, 58), which joins the
points of intersection, is a chord common to the two circles; and, if a perpendicular be erected upon the middle of this chord, it must pass through each of the centres C and D (106). But through two given points only one straight line can be drawn; therefore the straight line, which passes through the centres, will be perpendicular to the middle of the common chord.
114. If the distance of two centres is less than the sumn 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.
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.
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.
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. i. 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 he taken equal to DCE; and we have, according to what has just been demonstrated, AI = DE. But, by hypothesis, the arc AB - DE; we should consequently have AI=AB, or the part equal to the whole, which is impossible; therefore the angle ACB=DCE.
THEOREM. 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 num
bers, 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., D x, 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