2. Let the square A B C D (in the first figure) be described; and with the radii A B, B C, and D C, the three arches B D, C A, and A C; of which let the two B D and C A cut one another in E, and the two B D and A C in F. The diagonals therefore B D and A C being drawn will cut one another in the centre of the square G, and the two arches B D and C A in two equal parts in H and Y; and the arch B H D will be trisected in F and E. Through the centre G let the two strait lines K G L and M G N be drawn parallel and equal to the sides of the square A B and A D, cutting the four sides of the same square in the points K, L, M, and N; which being done, K L will pass through F, and M N through E. Then let O P be drawn parallel and equal to the side B C, cutting the arch B F D in F, and the sides A B and D C in O and P. Therefore O F will be the sine of the arch B F, which is an arch of 30 degrees; and the same O F will be equal to half the radius. Lastly, dividing the arch B F in the middle in Q, let R Q, the sine of the arch B Q, be drawn and produced to S, so that Q S be equal to R Q, and consequently R S be equal to the chord of the arch B F; and let F S be drawn and produced to T in the side B C. I say, the strait line B T is equal to the arch B F; and consequently that B V, the triple of B T, is equal to the arch of the quadrant B F E D. Let T F be produced till it meet the side B A produced in X; and dividing O F in the middle in Z, let Q Z be drawn and produced till it meet with the side B A produced. Seeing therefore the strait lines R S and O F are parallel, and divided in the midst in Q and Z, Q Z produced will fall upon X, and X Z Q produced to the side B C will cut B T in the midst in α.