101. A straight line cannot meet the circumference of a circle in more than two points.

For, if it could meet it in three, those three points would be equally distant from the centre; and hence, there would be three equal straight lines drawn from the same point to the same straight line, which is impossible (54.).

THEOREM.

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

micircle AMDB may be applied exactly to the semicircle ENGF, and the curve line AMDB will coincide entirely with the curve line ENGF. But the part AMD is equal to the part ENG (Hyp.); hence the point D will fall on G; therefore the chord AD is equal to the chord EG.

Conversely, supposing again the radii AC, EO, to be equal, if the chord AD is equal to the chord EG, the arcs AMD, ENG will be equal.

For, if the radii CD, OG, be drawn, the triangles ACD, EOG, having all their sides respectively equal, namely, AC= EO, CD=OG, and AD=EG, are themselves equal; and, consequently, the angle ACD is equal EOG. Now, placing the semicircle ADB on its equal EGF, since the angles ACD, EOG, are equal, it is plain that the radius CD will fall on the radius OG, and the point D on the point G; therefore the arc AMD is equal to the arc ENG.

103. In the same circle, or in equal circles, a greater arc is subtended by a greater chord, and conversely: the arcs being always supposed to be less than a semicircumference.

Let the arc AH be greater than AD (see the preceding figure); and draw the chords AD, AH, and the radii CD, CH. The two sides AC, CH, of the triangle ACH are equal to the two AC, CD, of the triangle ACD, and the angle ACH is greater than ACD; hence (42.) the third side AH is .greater than the third AD; therefore the chord, which subtends the greater arc, is the greater.

Conversely, if the chord AH is greater than AD, it will follow, on comparing the same triangles, that the angle ACH is greater than ACD; and hence, that the arc AH is greater than AD.

104. Scholium. The arcs here treated of are each less than the semicircumference. If they were greater, the reverse property would have place; as the arcs increased, the chords would diminish, and conversely. Thus, the arc AKBD being greater than AKBH, the chord AD of the first is less than the chord AH of the second.

105. The radius perpendicular to a chord, bisects it, and bisects also the subtended arc of the chord.

Let AB be a chord and CG the radius perpendicular to it; then AD= DB and the arc AG=GB.

Draw the radii CA, CB. These radii considered with regard to the perpendicular CD, are two equal oblique lines; hence (52.) they lie equally distant from that perpendicular: hence AD is equal to DB.

Again, since AD, DB, are equal, CG

H

B

is a perpendicular erected from the middle of AB; hence (55.) every point of this perpendicular must be equally distant from its two extremities A and B. Now, G is one of those points; therefore AG, BG, are equal. But, if the chord AG is equal to the chord GB, the arc AG will be equal to

the arc GB (102); hence, the radius CG, at right angles to the chord AB, divides the arc subtended by that chord into two equal parts at the point G.

106. Scholium. The centre C, the middle point D, of the chord AB, and the middle point G, of the arc subtended by this chord, are three points situated in the same line perpendicular to the chord. But two points are sufficient to determine the position of a straight line; hence every straight line which passes through two of the points just mentioned, will necessarily pass through the third, and be perpendicular to the chord.

It follows, likewise, that the perpendicular, raised from the middle of a chord passes through the centre, and through the middle of the arc subtended by that chord.

For this perpendicular is the same as the one let fall from the centre on the same chord, since both of them pass through the middle of the chord.

THEOREM.

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

Let A, B, and C, be the given points.

Join AB, BC; and bisect those straight lines by the perpendiculars DE, FG: we assert first, that DE and FG, will meet in some point O.

A

D

C

K

For, they must necessarily cut each other, if they are not parallel. Now, suppose they were parallel, the line AB, which is perpendicular to DE, would also be perpendicular to FG (65.); and the angle K would be a right angle; but BK, the production of BD, is different from BF, because the three points A, B, C, are not in the same straight line; hence there would be two perpendiculars, BF, BK, let fall from the same point on the same straight line, which (50.) is impossible; hence DE, FG, will always meet in some point O.

And moreover, this point O, since it lies in the perpendicular DE, is equally distant from the two points, A and B (55.); and since the same point O lies in the perpendicular FG, it is also equally distant from the two points B and C : hence the three distances OA, OB, OC, are equal; therefore

2

the circumference described from the centre O, with the radius OB, will pass through the three given points A, B, C.

We have now shewn that one circumference can always be made to pass through three given points, not in the same straight line we assert farther, that but one can described through them.

For, if there were a second circumference passing through the three given points A, B, C, its centre could not be out of the line DE (55.), for then it would be unequally distant from A and B; neither could it be out of the line FG, for a like reason; therefore, it would be in both the lines DE, FG. But two straight lines cannot cut each other in more than one point; hence there is but one circumference which can pass through three given points.

108. Cor. Two circumferences cannot meet in more than two points; for, if they have three common points, they must have the same centre, and form one and the same circumference.

109.

Two equal chords are equally distant from the centre; and of two unequal chords, the less is at the greater distance from

the centre.

First. Suppose the chord AB= DE. Bisect those chords by the perpendiculars CF, CG, and draw the radii CA, CD.

In the right-angled trianglesCAF, DCG, the hypotenuses CA, CD, are equal; and the side AF, the half of AB, is equal to the side DG, the half of DE: hence the triangles are equal (59.), and CF is equal to

CG; hence (first) the two equal chords AB, DE, are equally distant from the centre.

The

Secondly. Let the chord AH be greater than DE. arc AKH (103.) will be greater than DME: cut off from the former, a part equal to the latter, ANB=DM; draw the chord AB, and let fall CF perpendicular to this chord, and CI perpendicular to AH. It is evident that CF is greater than CO, and CO than CI (52.); therefore, CF is still greater than CI. But CF is equal to CG, be

cause the chords AB, DE, are equal; hence we have CG/CI; hence of two unequal chords, the less is the farther from the centre.

110. A straight line, perpendicular to a radius, at its extremity, is a tangent to the circumference.

without the circle; therefore, BD has no point but A common to it and the circumference; consequently BD (97.) is a tangent.

111. Scholium. From a given point A, only one tangent AD can be drawn to the circumference: for if another could be drawn, it would not be perpendicular to the radius CA; hence, in reference to this new tangent, the radius AC would be an oblique line, and the perpendicular let fall from the centre upon this tangent would be shorter than CA; hence this supposed tangent would enter the circle, and⚫ be a secant.

and the arc NH-HQ; and therefore MH-NH=HP-HQ; in other words, MN=PQ.