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.