For seeing the radius is a mean proportional between the chord of a quadrant and its semichord, and the same radius a mean proportional between the arch of the quadrant and Z, the square of the radius will be equal as well to the rectangle made of the chord and semichord of the quadrant, as to the rectangle made of the arch of the quadrant and Z; and these two rectangles will be equal to one another. Wherefore, as the arch of a quadrant is to its chord, so reciprocally is half the chord of the quadrant to Z. But as the arch of the quadrant is to its chord, so is half the arch of the quadrant to half the chord of the quadrant. Wherefore, as half the arch of the quadrant is to half the chord of the quadrant (or to the sine of 45 degrees), so is half the chord of the quadrant to Z; which was to be proved.
Lemma II. The radius, the arch of the semiquadrant, the sine of 45 degrees, and the semiradius, are proportional.
For seeing the sine of 45 degrees is a mean proportional between the radius and the semiradius; and the same sine of 45 degrees is also a mean proportional (by the [precedent lemma]) between the arch of 45 degrees and Z; the square of the sine of 45 degrees will be equal as well to the rectangle made of the radius and semiradius, as to the rectangle made of the arch of 45 degrees and Z. Wherefore, as the radius is to the arch of 45 degrees, so reciprocally is Z to the semiradius; which was to be demonstrated.
Let now A B C D (in [fig. 4]) be a square; and with the radii A B, B C and D A, let the three quadrants A B D, B C A and D A C, be described; and let the strait lines E F and G H, drawn parallel to the sides B C and A B, divide the square A B C D into four equal squares. They will therefore cut the arch of the quadrant A B D into three equal parts in I and K, and the arch of the quadrant B C A into three equal parts in K and L. Also let the diagonals A C and B D be drawn, cutting the arches B I D and A L C in M and N. Then upon the centre H with the radius H F equal to half the chord of the arch B M D, or to the sine of 45 degrees, let the arch F O be drawn cutting the arch C K in O; and let A O be drawn and produced till it meet with B C produced in P; also let it cut the arch B M D in Q, and the strait line D C in R. If now the strait line H Q be equal to the strait line D R, and being produced to D C in S, cut off D S equal to half the strait line B P; I say then the strait line B P will be equal to the arch B M D.
For seeing P B A and A D R are like triangles, it will be as P B to the radius B A or A D, so A D to D R; and therefore as well P B, A D and D R, as P B, A D (or A Q) and Q H are in continual proportion; and producing H O to D C in T, D T will be equal to the sine of 45 degrees, as shall by and by be demonstrated. Now D S, D T and D R are in continual proportion by the [first lemma]; and by the [second lemma] D C. D S:: D R. D F are proportionals. And thus it will be, whether B P be equal or not equal to the arch of the quadrant B M D. But if they be equal, it will then be, as that part of the arch B M D which is equal to the radius, is to the remainder of the same arch B M D; so A Q to H Q, or so B C to C P. And then will B P and the arch B M D be equal. But it is not demonstrated that the strait lines H Q and D R are equal; though if from the point B there be drawn (by the construction of [fig. 1]) a strait line equal to the arch B M D, then D R to H Q, and also the half of the strait line B P to D S, will always be so equal, that no inequality can be discovered between them. I will therefore leave this to be further searched into. For though it be almost out of doubt, that the strait line B P and the arch B M D are equal, yet that may not be received without demonstration; and means of demonstration the circular line admitteth none that is not grounded upon the nature of flexion, or of angles. But by that way I have already exhibited a strait line equal to the arch of a quadrant in the first and second aggression.
It remains that I prove D T to be equal to the sine of 45 degrees.
In B A produced let A V be taken equal to the sine of 45 degrees; and drawing and producing V H, it will cut the arch of the quadrant C N A in the midst in N, and the same arch again in O, and the strait line D C in T, so that D T will be equal to the sine of 45 degrees, or to the strait line A V; also the strait line V H will be equal to the strait line H I, or the sine of 60 degrees.
For the square of A V is equal to two squares of the semiradius; and consequently the square of V H is equal to three squares of the semiradius. But H I is a mean proportional between the semiradius and three semiradii; and, therefore, the square of H I is equal to three squares of the semiradius. Wherefore H I is equal to H V. But because A D is cut in the midst in H, therefore V H and H T are equal; and, therefore, also D T is equal to the sine of 45 degrees. In the radius B A let B X be taken equal to the sine of 45 degrees; for so V X will be equal to the radius; and it will be as V A to A H the semiradius, so V X the radius to X N the sine of 45 degrees. Wherefore V H produced passes through N. Lastly, upon the centre V with the radius V A let the arch of a circle be drawn cutting V H in Y; which being done, V Y will be equal to H O (for H O is, by construction, equal to the sine of 45 degrees) and Y H will be equal to O T; and, therefore, V T passes through O. All which was to be demonstrated.
I will here add certain problems, of which if any analyst can make the construction, he will thereby be able to judge clearly of what I have now said concerning the dimension of a circle. Now these problems are nothing else (at least to sense) but certain symptoms accompanying the construction of the [first] and [third figure] of this chapter.
Describing, therefore, again, the square A B C D (in [fig. 5]) and the three quadrants A B D, B C A and D A C, let the diagonals A C and B D be drawn, cutting the arches B H D and C I A in the middle in H and I; and the strait lines E F and G L, dividing the square A B C D into four equal squares, and trisecting the arches B H D and C I A, namely, B H D in K and M, and C I A in M and O. Then dividing the arch B K in the midst in P, let Q P the sine of the arch B P, be drawn and produced to R, so that Q R be double to Q P; and, connecting K R, let it be produced one way to B C in S, and the other way to B A produced in T. Also let B V be made triple to B S, and consequently, (by the [second article] of this chapter) equal to the arch B D. This construction is the same with that of the first figure, which I thought fit to renew discharged of all lines but such as are necessary for my present purpose.