tangent can be arawn, for two lines could not be both perpendicular to the diameter at the same point.

Cor. 2. Tangents at each extremity of a diameter are parallel. (Prop. XII. Cor. 5. B. I.)

Cor. 3. And conversely, parallel tangents are both perpendicular to the same diameter (Prop. XIV. Cor. 5. B. I.), and have their points of contact at its extremities.

Scholium.

This proposition shows the possibility of the existence of a tangent to a circle (Post. IV. B. I.).

PROPOSITION X. THEOREM.

The arcs of a circle intercepted by two parallels are equal.

If the parallels are tangents to the circle, as AB, CD, then each of the arcs intercepted is a semi-circumference, as their points of contact coincide with the extremities of the diameter EF (Prop. IX. Cor. 3.).

Next, let both the parallel lines be chords, as GH, JK, and let the diameter FE be perpendicular to one of them, as GH: it will also be perpendicular to the other, since they are parallel; therefore this diameter must bisect each of the arcs which they subtend, that is, GE is equal to EH, and JE to EK; therefore JE diminished by GE, is equal to EK diminished by EH; that is, in this case likewise, the intercepted arcs JG, KH, are equal.

PROPOSITION XI.

THEOREM.

(Converse of Prop. X.) If two straight lines intercept equal arcs of a circle, and do not cut each other within the circle, the lines will be parallel.

If the two lines be AB, CD (preceding diagram), which touch the circumference, and if, at the same time, the intercepted arcs EJF, EKF are equal, EF must be a diameter

(Prop. I.); and therefore AB, CD, are parallel (Prop. IX. Cor. 2.)

But if only one of the lines, as AB, touch, while the other, GH, cuts the circumference, making the arcs EG, EH equal, then the diameter FE, which bisects the arc GEH, is perpendicular to its chord GH (Prop. V. Cor. 1.): it is also perpendicular to the tangent AB (Prop. IX.); therefore AB, GH are parallel.

If both lines cut the circle, as GH, JK, and intercept equal arcs GJ, HK, let the diameter FE bisect one of the chords, as GH: it will also bisect the arc GEH (Prop. V.), so that EG is equal to EH; and since GJ is by hypothesis equal to HK, the whole arc EJ is equal to the whole arc EK; therefore the chord JK is bisected by the diameter FE: as therefore both chords are bisected by this diameter, they are perpendicular to it, (Prop. V. Cor. 1.) that is, they are parallel.

Scholium.

The restriction in the enunciation of this proposition, viz. that the lines do not cut each other within the circle, is necessary; for lines drawn through the points G, K and J, H will intercept the equal arcs GJ, HK, and yet not be parallel, since they will intersect each other within the circle.

If two circumferences have a point in common, situated in the straight line joining their centres, the circumferences will touch each other in that point. For if they have also another point in common, then a line joining the two points will lie wholly within each circle (Prop. II.), and therefore the two lines from the middle of this chord to the centre of each circle must both be perpendicular to it (Prop. V. Cor. 1), and must therefore form but one straight line; but by hypothesis, the centres are in the line passing through the other point, so that two distinct lines have the same extremities, which is impossible.

Cor. 1. If two circumferences cut each other, the points of

intersection must both be without the line joining the centres; for if either were in this line, the circumferences would touch. Cor. 2. If the distance between the centres of two circles be equal to the sum of their radii, the circumferences touch externally.

Cor. 3. But if the distance be equal to the difference of the radii, one touches the other internally, for in both cases the circumferences pass through the same point in the line joining the

centres.

Scholium.

Hence the possibility of one circle touching another (Post. B. III.).

If two circumferences have a point in common, situated out of the line joining their centres, the circumferences will cut each other.

Let C and D be the centres of two circles, and let A be a point situated out of the line CD, common to the two circumferences.

Let AB be a perpendicular to CD, then AB, if produced, shall again cut the circumferences: for if it were a tangent to either, it would be perpendicular either to DA, or CA, which it is not; let it then cut the circumference, whose centre is C, in the point E, then the chord AE is bisected in B (Prop. V.); let it cut the other circle in F, then AF is also bisected in B; consequently the points E and F coincide, that is, the circumferences again have a point in common.

E

Cor. 1. Hence the converse of proposition XII. viz. if two circumferences touch each other, their centres and point of contact lie in the same straight line; for if the point of contact lay out of the line joining the centres, the circles would cut.

Cor. 2. Therefore, if two circumferences touch each other, the distance of their centres is equal either to the sum or difference of their radii, accordingly as they touch externally or

internally, which is the converse of Corollaries 2 and 3, last proposition.

Cor. 3. It moreover follows from the above demonstration, that the line joining the intersections of the circumferences, is bisected at right angles by the line joining the centres; for it is shown that the perpendicular to this line, from one of the points, passes through the other, and is bisected by the line joining the centres.

Scholium.

1. Corollary 1, to proposition XII., proves the converse of this proposition.

2. A mere inspection of the preceding diagrams shows, that if two circumferences cut, the distance of the centres must be less than the sum of the radii; for CD is less than the sum of CA, DA (Prop. XXI. B. I.); and, consequently, that, if the distance of the centres of two circles be greater than the sum of their radii, the circumferences will neither touch (Prop. XIIÏ. Cor. 2.) nor cut.

3. It is, moreover, equally plain that, if two circumferences cut, the distance of the centres must exceed the difference of the radii; for CD is longer than the difference of CA, DA (Prop. XXI. Cor. B. I.): consequently, if the distance of the centres of two circles be less than the difference of their radii, their circumferences will neither touch (Cor. 2.) nor cut.

4. It appears, therefore, that in order that two circumferences may cut, the distance of their centres must be less than the sum, and greater than the difference of the radii.

An inscribed angle is equal to an angle at the centre of the circle whose sides include half the arc included by those of the inscribed angle.

Let ADB be an inscribed angle, the sides of which include the arc AB; this angle is equal to one at the centre C, whose sides include half that arc.

Suppose, first, that the centre lies in one of the sides DB of the proposed angle, let CA be drawn, as also CE, bisecting the angle ACB.

Then, since ACD is an isosceles triangle, the exterior angle ACB is double the interior opposite angle D; therefore the half thereof ACE is equal to D, and its sides intercept half the arc AB (Prop. III. Cor. 2.).

E

B

E

Next, let the centre C lie within the sides of the angle ADB, and let DCE be drawn, as also CF to the middle of the arc AE, and CG to the middle of the arc BE.

Then, by the preceding case, the angle ADE is equal to the angle FCE, and the angle EDB to the angle ECG; hence, the whole angle ADB is equal to the

whole angle FCG; and the arc FG, subtended by the angle at the centre, is, by construction, equal to half the arc AEB, subtended by that at the circumference.

Lastly, let the centre lie without the sides of the angle ADB, and draw, as before, DE through the centre C; CF to the middle of the arc AE, and CG to the middle of the arc EB.

D

Α

B

Then, by the first case, the angle ADE is equal to the angle FCE, and EDB to ECG; hence the angle ADB, the difference of the angles at the circumference is equal to FCG, the difference of the angles at the centre. Now, the arc FG is, by construction, equal to the difference of half the arcs EB, EA, or, which is the same thing, to half the difference of those arcs, that is, to half the arc AB; so that in every case the inscribed angle is equal to an angle at the centre, whose sides intercept half the arc included by those of the former.

Cor. 1. Hence an angle at the centre of a circle is double an angle at the circumference, subtending the same arc.

Cor. 2. Angles in the same, or in equal segments, or, in other words, inscribed angles, subtended by the same, or equal arcs, are equal; each being equal to the angle at the centre, whose sides include half the equal arcs.

Cor. 3. An angle in a semi-circle is a right angle; for it is equal to an angle at the centre, subtended by half a semi-circumference (Prop. IV. Cor. 3.).

Cor. 4. If, in a circle, two chords drawn from a point in the circumference be respectively equal to two chords drawn from another point, they shall include equal angles; for the equal chords subtending equal arcs (Prop. IV. Cor. 2.), each angle must include the same portion of the circumference.