Page images

If the radii AC, EO, are equal, and also the arcs AMD, ENG; then the chord AD will be equal to the A chord EG.

For, since the diameters AB, EF, are equal, the semicircle 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, by hypothesis; 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 also be equal.

For, if the radii CD, OG, be drawn, the triangles ACD, EOG, will have all their sides equal, each to each, namely, AC=EO, CD=OG, and AD=EG; hence the triangles are themselves equal; and, consequently, the angle ACD is equal EOG (Book I. Prop. X.). 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.


In the same circle, or in equal circles, a greater arc is subtended by a greater chord, and conversely, the greater chord subtends the greater arc.

Let the arc AH be greater than the arc AD; then will the chord AH be greater than the chord AD.




For, draw 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, the third side AH is greater than the third side AD (Book I. Prop. IX.); 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.

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


The radius which is perpendicular to a chord, bisects the chord, and bisects also the subtended arc of the chord.

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

For, draw the radii CA, CB. Then the two right angled triangles ADC, CDB, will have AČ=CB, and CD common; hence, AD is equal to DB (Book I. Prop. XVII.).

Again, since AD, DB, are equal, CG is a perpendicular erected from the mid



dle of AB; hence every point of this perpendicular must be equally distant from its two extremities A and B (Book I. Prop. XVI.). Now, G is one of these 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 (Prop. IV.); 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.

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 of 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 througli 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 of the circle, 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 centre and middle of the chord.


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.

Draw AB, BC, and bisect these straight lines by the perpendiculars DE, FG: we say first, that DE and FG, will meet in some point O.

For, they must necessarily cut each other, if they are not parallel.


Now, if they were parallel, the line AB, which is perpendicular to DE, would also be perpendicular to FG, and the angle K would be a right angle (Book I. Prop. XX. Cor. 1.). But BK, the prolongation of BD, is a different line 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 B, on the same straight line, which is impossible (Book I. Prop. XIV.); 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 (Book I. Prop. XVI.); 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 the circumference described from the centre O, with the radius OB, will pass through the three given points A, B, C.

We have now shown that one circumference can always be made to pass through three given points, not in the same straight line: we say farther, that but one can be 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, for then it would be unequally distant from A and B (Book I. Prop. XVI.); 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.

Cor. Two circumferences cannot meet in more than two points; for, if they have three common points, there would be two circumferences passing through the same three points; which has been shown by the proposition to be impossible.


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

[merged small][ocr errors][merged small][merged small][merged small]

In the right angled triangles CAF, DCG, the hypothenuses 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, and CF is equal to CG (Book I. Prop. XVII.); hence, the two equal chords AB, DE, are equally distant from the centre.



Secondly Let the chord AH be greater than DE. The arc AKH will be greater than DME (Prop. V.): cut off from the former, a part ANB, equal to DME; 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 (Book I. Prop. XV.); therefore, CF is still greater than CI. But CF is equal to CG, because the chords AB, DE, are equal: hence we have CG>CI; hence of two unequal chords, the less is the farther from the centre.


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

Let BD be perpendicular to the B radius CA, at its extremity A; then will it be tangent to the circumfe


For, every oblique line CE, is longer than the perpendicular CA (Book I. Prop. XV.); hence the


point E is without the circle; therefore, BD has no point but A common to it and the circumference; consequently BD is a tangent (Def. 8.).

Scholium. At 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 (Book I. Prop. XIV. Sch.); 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



Two parallels intercept equal arcs on the circumference.

There may be three cases.

First. If the two parallels are secants, draw the radius CH perpendicular to the chord MP. It will, at the same time be perpendicular to NQ (Book J.Prop.XX.Cor.1.); therefore, the point H will be at once the middle of the arc MHP, and of the arc NHQ (Prop. VI.); therefore, we shall have the arc MH-HP, and the arc NH

[blocks in formation]

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


Second. When, of the two parallels AB, DE, one is a secant, the other a tangent, draw the radius CH Ato the point of contact H; it will be perpendicular to the tangent DE (Prop. IX.), and also to its parallel MP. But, since CH is perpendicular to the chord MP, the point H must be the middle of the arc MHP (Prop. VI.); therefore the arcs MH, HP, included between the parallels AB, DE, are equal.


[blocks in formation]


[blocks in formation]

Third. If the two parallels DE, IL, are tangents, the one at H, the other at K, draw the parallel secant AB; and, from what has just been shown, we shall have MH-HP, MK=KP; and hence the whole arc HMK=HPK. It is farther evident that each of these arcs is a semicircumference.

« PreviousContinue »