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Fig. 50.

Fig. 50.

THEOREM.

102. In the same circle, or in equal circles, equal arcs are subtended by equal chords, and conversely, equal chords subtend equal

arcs.

Demonstration. The radius AC (fig. 50) being equal to the radius EO, and the arc AMD equal to the arc ENG; the chord AD will be equal to the chord EG.

For, the diameter AB being equal to the diameter EF, the semicircle AMDB may be applied exactly to the semicircle ENGF, and then the curved line AMDB will coincide entirely with the curved line ENGF; but the portion AMD being supposed equal to the portion ENG, the point D will fall upon G; therefore the chord AD is equal to the chord EG.

Conversely, AC being supposed equal to EO, if the chord AD= EG, the arc AMD will be equal to the arc ENG.

For, if the radii CD, OG, be drawn, the two triangles ACD, EOG, will have the three sides of the one equal to the three sides of the other, each to each, namely, AC EO, CD = OG and AD = EG; therefore these triangles are equal (43); hence the angle ACD=EOG. Now, if the semicircle ADB be placed upon EGF, because the angle ACD = EOG, it is evident, that the radius CD will fall upon the radius OG, and the point D upon G, therefore the arc AMD is equal to the arc ENG.

THEOREM.

103. In the same circle, or in equal circles, if the arc be less than half a circumference, the greater arc is subtended by the greater chord; and, conversely, the greater chord is subtended by the greater

arc.

Demonstration. Let the arc AH (fig. 50) be greater than AD, and let the chords AD and AH, and the radii CD, CH, be drawn. The two sides, AC, CH, of the triangle ACH, are equal to the two sides AC, CD, of the triangle ACD, and the angle ACH is greater than ACD; hence the third side AH is greater than the third side AD (42), therefore the greater arc is subtended by the greater chord.

Conversely, if the chord AH be greater than AD, it may be inferred from the same triangles that the angle ACH is greater than ACD. and that thus the arc AH is greater than AD.

104. Scholium. The arcs of which we have been speaking, are supposed to be less than a semicircumference; if they were greater, the contrary would be true; in this case, as the arc increases, the chord would diminish, and the reverse; thus, the arc AKBD being greater than AKBH, the chord AD of the first is less than the chord AH of the second.

THEOREM.

105. The radius CG (fig. 51), perpendicular to a chord AB, Fig. 51. bisects this chord and the arc subtended by it AGB.

Demonstration. Draw the radii CA, CB; these radii are, with respect to the perpendicular CD, two equal oblique lines, therefore they are equally distant from the perpendicular (52), and AD= DB.

Again, since AD = BD, and CG is a perpendicular erected upon the middle of AB, each point in CG is at equal distances from A and B (55). The point G is one of these points; therefore AG=GB. But, if the chord AG is equal to the chord GB, the arc AG will be equal to the arc GB (102); therefore the radius CG, perpendicular to the chord AB, bisects the arc subtended by this chord in the point G.

106. Scholium. The centre C, the middle D of the chord AB, and the middle G of the arc subtended by this chord, are three points situated in the same straight line perpendicular to the chord. Now, two points in a straight line are sufficient to determine its position; therefore a straight line which passes through any two of these points must necessarily pass through the third; and must be perpendicular to the chord.

It follows also, that a perpendicular erected upon the middle of a chord passes through the centre, and the middle of the arc subtended by that chord.

For this perpendicular is the same as that let fall from the centre upon the same chord, since they both pass through the middle of the chord (51). ~

THEOREM.

107. The circumference of a circle may be made to pass through any three points, A, B, C (fig. 52), which are not in the same Fig 32: Geom.

4

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 produced, 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 impossible (50); therefore the perpendiculars DE, FG, will always cut each other in some point O.

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 OA, 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 each 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.

THEOREM.

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 ANBDME, 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 = CG, 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.

THEOREM.

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 draw 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.

Fig. 55.

Fig. 56.

THEOREM.

112. Two parallels AB, DE (fig. 55), intercept upon the circumference equal arcs MN, PQ.

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 MHHP, and the arc NH = HQ; also

MH-NH HP-HQ, that is. MN PQ.

==

=

2. If of the two parallels AB, DE (fig. 56), one be a secant and the other a tangent; to the point of contact H draw the radius 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 semicircumfer

ence.

=

Fig. 57,

58.

THEOREM.

113. If the circumferences of two circles cut 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.

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.

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