For let A B and A C (in the [second figure]) be two circles, touching one another, and the strait line A D in the point A; and let their centres be E and F; and let it be supposed, that as A E is to A F, so is the arch A B to the arch A H. I say the crookedness of the arch A C is to the crookedness of the arch A H, as A E is to A F. For let the strait line A D be supposed to be equal to the arch A B, and the strait line A G to the arch A C; and let A D, for example, be double to A G. Therefore, by reason of the likeness of the arches A B and A C, the strait line A B will be double to the strait line A C, and the radius A E double to the radius A F, and the arch A B double to the arch A H. And because the strait line A D is so bowed to be coincident with the arch A B equal to it, as the strait line A G is bowed to be coincident with the arch A C equal also to it, the flexion of the strait line A G into the crooked line A C will be equal to the flexion of the strait line A D into the crooked line A B. But the flexion of the strait line A D into the crooked line A B is double to the flexion of the strait line A G into the crooked line A H; and therefore the flexion of the strait line A G into the crooked line A C is double to the flexion of the same strait line A G into the crooked line A H. Wherefore, as the arch A B is to the arch A C or A H; or as the radius A E is to the radius A F; or as the chord A B is to the chord A C; so reciprocally is the flexion or uniform crookedness of the arch A C, to the flexion or uniform crookedness of the arch A H, namely, here double. And this may by the same method be demonstrated in circles whose perimeters are to one another triple, quadruple, or in whatsoever given proportion. The crookedness therefore of two equal arches taken in several circles are in proportion reciprocal to that of their radii, or like arches, or like chords; which was to be demonstrated.
Let the square A B C D be again described (in the [third figure]), and in it the quadrants A B D, B C A and D A C; and dividing each side of the square A B C D in the midst in E, F, G and H, let E G and F H be connected, which will cut one another in the centre of the square at I, and divide the arch of the quadrant A B D into three equal parts in K and L. Also the diagonals A C and B D being drawn will cut one another in I, and divide the arches B K D and C L A into two equal parts in M and N. Then with the radius B F let the arch F E be drawn, cutting the diagonal B D in O; and dividing the arch B M in the midst in P, let the strait line E a equal to the chord B P be set off from the point E in the arch E F, and let the arch a b be taken equal to the arch O a, and let B a and B b be drawn and produced to the arch A N in c and d; and lastly, let the strait line A d be drawn. I say the strait line A d is equal to the arch A N or B M.
I have proved in the preceding article, that the arch E O is twice as crooked as the arch B P, that is to say, that the arch E O is so much more crooked than the arch B P, as the arch B P is more crooked than the strait line E a. The crookedness therefore of the chord E a, of the arch B P, and of the arch E O, are as 0, 1, 2. Also the difference between the arches E O and E O, the difference between the arches E O and E a, and the difference between the arches E O and E b, are as 0, 1, 2. So also the difference between the arches A N and A N, the difference between the arches A N and A c, and the difference between the arches A N and A d, are as 0, 1, 2; and the strait line A c is double to the chord B P or E a, and the strait line A d double to the chord E b.
Again, let the strait line B F be divided in the midst in Q, and the arch B P in the midst in R; and describing the quadrant B Q S (whose arch Q S is a fourth part of the arch of the quadrant B M D, as the arch B R is a fourth part of the arch B M, which is the arch of the semiquadrant A B M) let the chord S e equal to the chord B R be set off from the point S in the arch S Q; and let B e be drawn and produced to the arch A N in f; which being done, the strait line A f will be quadruple to the chord B R or S e. And seeing the crookedness of the arch S e, or of the arch A c, is double to the crookedness of the arch B R, the excess of the crookedness of the arch A f above the crookedness of the arch A c will be subduple to the excess of the crookedness of the arch A c above the crookedness of the arch A N; and therefore the arch N c will be double to the arch c f. Wherefore the arch c d is divided in the midst in f, and the arch N f is ¾ of the arch N d. And in like manner if the arch B R be bisected in V, and the strait line B Q in X, and the quadrant B X Y be described, and the strait line Y g equal to the chord B V be set off from the point Y in the arch Y X, it may be demonstrated that the strait line B g being drawn and produced to the arch A N, will cut the arch f d into two equal parts, and that a strait line drawn from A to the point of that section, will be equal to eight chords of the arch B V, and so on perpetually; and consequently, that the strait line A d is equal to so many equal chords of equal parts of the arch B M, as may be made by infinite bisections. Wherefore the strait line A d is equal to the arch B M or A N, that is, to half the arch of the quadrant A B D or B C A.
Coroll. An arch being given not greater than the arch of a quadrant (for being made greater, it comes again towards the radius B A produced, from which it receded before) if a strait line double to the chord of half the given arch be adapted from the beginning of the arch, and by how much the arch that is subtended by it is greater than the given arch, by so much a greater arch be subtended by another strait line, this strait line shall be equal to the first given arch.
Supposing the strait line B V (in [fig. 1]) be equal to the arch of the quadrant B H D, and A V be connected cutting the arch B H D in I, it may be asked what proportion the arch B I has to the arch I D. Let therefore the arch A Y be divided in the midst in o, and in the strait line A D let A p be taken equal, and A q double to the drawn chord A o. Then upon the centre A, with the radius A q, let an arch of a circle be drawn cutting the arch A Y in r, and let the arch Y r be doubled at t; which being done, the drawn strait line A t (by what has been last demonstrated) will be equal to the arch A Y. Again, upon the centre A with the radius A t let the arch t u be drawn cutting A D in u; and the strait line A u will be equal to the arch A Y. From the point u let the strait line u s be drawn equal and parallel to the strait line A B, cutting M N in x, and bisected by M N in the same point x. Therefore the strait line A x being drawn and produced till it meet with B C produced in V, it will cut off B V double to B s, that is, equal to the arch B H D. Now let the point, where the strait line A V cuts the arch B H D, be I; and let the arch D I be divided in the midst in y; and in the strait line D C, let D z be taken equal, and D δ double to the drawn chord D y; and upon the centre D with the radius D δ let an arch of a circle be drawn cutting the arch B H D in the point n; and let the arch n m be taken equal to the arch I n; which being done, the strait line D m will (by the last foregoing corollary) be equal to the arch D I. If now the strait lines D m and C V be equal, the arch B I will be equal to the radius A B or B C; and consequently X C being drawn, will pass through the point I. Moreover, if the semicircle B H D ϐ being completed, the strait lines ϐ I and B I be drawn, making a right angle (in the semicircle) at I, and the arch B I be divided in the midst at i, it will follow that A i being connected will be parallel to the strait line ϐ I, and being produced to B C in k, will cut off the strait line B k equal to the strait line k I, and equal also to the strait line A γ cut off in A D by the strait line ϐ I. All which is manifest, supposing the arch B I and the radius B C to be equal.
But that the arch B I and the radius B C are precisely equal, cannot (how true soever it be) be demonstrated, unless that be first proved which is contained in [art. 1], namely, that the strait lines drawn from X through the equal parts of O F (produced to a certain length) cut off so many parts also in the tangent B C severally equal to the several arches cut off; which they do most exactly as far as B C in the tangent, and B I in the arch B E; insomuch that no inequality between the arch B I and the radius B C can be discovered either by the hand or by ratiocination. It is therefore to be further enquired, whether the strait line A V cut the arch of the quadrant in I in the same proportion as the point C divides the strait line B V, which is equal to the arch of the quadrant. But however this be, it has been demonstrated that the strait line B V is equal to the arch B H D.
The third attempt; and some things propounded to be further searched into.
4. I shall now attempt the same dimension of a circle another way, assuming the two following lemmas.
Lemma I. If to the arch of a quadrant, and the radius, there be taken in continual proportion a third line Z; then the arch of the semiquadrant, half the chord of the quadrant, and Z, will also be in continual proportion.