Scholium. At a given point A, only one tangent AD can be drawn to the circumference; for, if another could be drawr, 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

secant.

PROPOSITION X. THEOREM.

Two parallels intercept equal arcs on the circumference.

There may be three cases.

A/M

D/N

G

F

P\B

C+

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

D

M

F

P

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.

I

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.

PROPOSITION XI. THEOREM.

If 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 divide it into two equal parts.

For, let the line AB join the points of intersection. It will be a common chord to the two circles. Now if a perpendicular

B

be erected from the middle of this chord, it will pass through each of the two centres C and D (Prop. VI. Sch.). But no more than one straight line can be drawn through two points; hence the straight line, which passes through the centres, will bisect the chord at right angles.

PROPOSITION XII. THEOREM.

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

For, to make an intersection possible, the triangle CAD must be possible. Hence, not only must we have CD<AC+AD, but also the greater radius AD< AC+CD (Book I. Prop. VII.). And, whenever the triangle CAD can be constructed, it is plain

that the circles described from the centres C and D, will cut each other in A and B.

E

PROPOSITION XIII. THEOREM.

If the distance between the centres of two circles is equal to the sum of their radii, the two circles will touch each other externally.

Let C and D be the centres at a distance from each other equal to CA+AD.

The circles will evidently have the point A common, and they will have no other; because, if they had two points common, the distance between

C

their centres must be less than the sum of their radii.

X

PROPOSITION XIV. THEOREM.

If the distance between the centres of two circles is equal to the difference of their radii, the two circles will touch each other internally.

Let C and D be the centres at a dis- E tance from each other equal to AD-CA.

It is evident, as before, that they will have the point A common; they can have no other; because, if they had, the greater radius AD must be less than the sum of the radius AC and the distance CD between the centres (Prop. XII.); which is contrary to the supposition.

Cor. Hence, if two circles touch each other, either externally or internally, their centres and the point of contact will be in the same right line.

Scholium. All circles which have their centres on the right line AD. and which pass through the point A, are tangent to each other. For, they have only the point A common, and if through the point A, AE be drawn perpendicular to AD, the straight line AE will be a common tangent to all the circles.

PROPOSITION XV. THEOREM.

In the same circle, or in equal circles, equal angles having their vertices at the centre, intercept equal arcs on the circumference: and conversely, if the arcs intercepted are equal, the angles contained by the radii will also be equal.

Let C and C be the centres of equal circles, and the angle ACB=DCE.

First. Since the angles ACB, DCE, are equal, they may be placed upon each other; and since their sides are equal, the point A will evidently fall on D, and the point B on E. But, in A that case, the arc AB must also

D

fall on the arc DE; for if the arcs did not exactly coincide, there would, in the one or the other, be points unequally distant from the centre; which is impossible hence the arc AB is equal to DE.

Secondly. If we suppose AB-DE, the angle ACB will be equal to DCE. For, if these angles are not equal, suppose ACB to be the greater, and let ACI be taken equal to DCE. From what has just been shown, we shall have AI=DE: but, by hypothesis, AB is equal to DE; hence AI must be equal to AB, or a part to the whole, which is absurd (Ax. 8.): hence, the angle ACB is equal to DCE.

PROPOSITION XVI. THEOREM.

In the same circle, or in equal circles, if two angles at the centre are to each other in the proportion of two whole numbers, the intercepted arcs will be to each other in the proportion of the same numbers, and we shall have the angle to the angle, as the corresponding arc to the corresponding arc.

Suppose, for example, that the angles ACB, DCE, are to each other as 7 is to 4; or, which is the same thing, suppose that the angle M, which may serve as a common measure, is contained times in the angle ACB, and 4 times in DCE

M

E

The seven partial angles ACm, mCn, nCp, &c., into which ACB is divided, being each equal to any of the four partial angles into which DCE is divided; each of the partial arcs Am, mn, np, &c., will be equal to each of the partial arcs Dr, xy, &c. (Prop. XV.). Therefore the whole arc AB will be to the whole arc DE, as 7 is to 4. But the same reasoning would evidently apply, if in place of 7 and 4 any numbers whatever were employed; hence, if the ratio of the angles ACB, DCE, can be expressed in whole numbers, the arcs AB, DE, will be to each other as the angles ACB, DCE.

Scholium. Conversely, if the arcs, AB, DE, are to each other as two whole numbers, the angles ACB, DCE will be to each other as the same whole numbers, and we shall have ACB DCE: AB: DE. For the partial arcs, Am, mn, &c. and Dx, xy, &c., being equal, the partial angles ACm, mCn, &c. and DCx, xCy, &c. will also be equal.

PROPOSITION XVII. THEOREM.

Whatever be the ratio of two angles, they will always be to each other as the arcs intercepted between their sides; the arcs being described from the vertices of the angles as centres with equal radii.

Let ACB be the greater and ACD the less angle. Let the less angle be placed on the greater. If the proposition is not truc, the angle ACB will be to the angle ACD as the arc AB is to an arc

greater or less than AD). Suppose this arc to be greater, and let it be represented by AO; we shall thus have, the angle ACE: angle ACD:: arc AB: arc AO. Next conceive the are