Secondly. Let the chord AH be greater than DE. The arc AKH (5. II.) will be greater than DME: cut off from the former, a part equal to the latter, ANB-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 (16. I.); 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.

PROPOSITION IX. THEOREM.

The straight BD, perpendicular to the radius CA, and passing through its extremity, is a tangent to the circumference.

FOR (16. I.) every oblique line B CE, is longer than the perpendicular CA; hence the point E is without the circle; hence BD has no point but A common to it with the circumference ; hence BD (Def. 8. II.) is a tangent.

A

E

D

Scholium. From a given point A, no more than 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.

PROPOSITION X. THEOREM.

Two parallels AB, DE, intercept equal arcs MN, PQ, 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 (23. I.); there- A/M fore, the point H (6. II.) will be at once the middle of the arc MHP, and of the arc NHQ; therefore, we shall have the arc MH=HP, and the arc NH=HQ; and therefore MH-NH-HP—HQ; in other words, MN-PQ.

E

PB

Second. When, of the two parallels D AB, DE, one is a secant, the other a tangent, draw the radius CH to the point of contact H; it will be perpendicular to the tangent DE (9. II.), 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; therefore the arcs I MH, HP, included between the parallels AB, DE, are equal.

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 shewn, 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 the line AB, which joins the points of intersection, is a chord common to the two circles. And if a perpendicular be

"

erected from the middle of this chord, it will pass (6. II.) through each of the two centres C and D. 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 circles will cut each other.

For to make an intersection possible, the triangle CAD (see the preceding figure) must be possible. Hence, not only must we have CDAC + AD, but also the greater radius ADAC + CD. 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.

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

THEY will evidently have the point A common, and they will have no other; because, if they had two points common, the distance between their centres must be less than the sum of their radii.

D

PROPOSITION XIV. THEOREM.

If the distance CD, between the centres of two circles, is equal to the difference of their radii CA, AD, those two circles will touch each other externally.

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 (12. II.) must be less than the sum of the radius AC and the distance between the centres, which it is not.

A

PROPOSITION XV. THEOREM.

In the same circle, or in equal circles, the equal angles ACB, DCE, having their vertices in the centre, intercept equal arcs AB, DE, on the circumference: and conversely, if the arcs AB, DE are equal, the angles ACB, DCE will also be equal.

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 that case, the arc AB must also

Q Q

fall on the arc DE; for if they 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 those angles are not equal, let ACB be the greater, and let ACI be taken equal to DCE. From what has just been shewn, we shall have AI-DE: but, by hypothesis, AB is equal to DE; hence AI must be equal to AB, the part to the whole, which is absurd: 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 ACB, DCE are to each other in the proportion of two whole numbers, the intercepted arcs AB, DE will be to each other in the proportion of the same numbers, and we shall have the angle ACB: angle DCE:: arc AB: arc DE.

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, whɔth may serve as a common measure, is

C

contained 7 times in the angle ACB and 4 times in DCE. 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 (15. II.) be equal to each of the partial arcs Ax, xy, &c. 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 DC x, xCy, &c. will also be equal.

PROPOSITION XVII. THEOREM.

Whatever be the ratio of the two angles ACB, ACD, those two angles will always be to each other as the arcs AB, AD intercepted between their sides, and described from their vertices as centres, with equal radii.

Let the less angle be placed on the greater. If the proposition is not true, 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 great

C

DIO

er, and let it be represented by AO; we shall thus have the angle ACB angle ACD:: arc AB: arc AO. Next conceive the arc AB to be divided into equal parts, each of which is less than DO; there will be at least one point of division between D and O; let I be that point; and join CI. The arcs AB, AI will be to each other as two whole numbers, and by the preceding theorem, we shall have the angle ACB: angle ACI :: arc AB: arc AI. Comparing these two proportions with each other, and observing that the antecedents are the same, we infer that the consequents are proportional, and thus we find the angle ACD: angle ACI :: arc AO: arc AI. But the arc AO is greater than the arc AI; hence, if this proportion is true, the angle ACD must be greater than the angle ACI on the contrary, however, it is less; hence the angle ACB cannot