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.