to the purpose; nor had it been though you had spoken more properly, and said _solent_, leaving out yourselves. [Illustration: _Six Lessons._ _Vol. VII. Eng. p.310_II. 325_ ] My first article hath this title, “_from a false supposition, a false quadrature of the circle_.” Seeing therefore you were resolved to show where I erred, you should have proved either that the supposition was true, and the conclusion falsely inferred, or contrarily, that though the supposition be false, yet the conclusion is true; for else you object nothing to my geometry, but only to my judgment, in thinking fit to publish it; which nevertheless you cannot justly do, seeing it was likely to give occasion to ingenious men (the practice of it being so accurate to sense) to inquire wherein the fallacy did consist. And for the problem as it was first printed, but never published, and consequently ought to have passed for a private paper stolen out of my study, your public objecting against it (in the opinion of all men that have conversed so much with honest company as to know what belongs to civil conversation), was sufficiently barbarous in divines. And seeing you knew I had rejected that proposition, it was but a poor ambition to take wing as you thought to do, like beetles from my egestions. But let that be as it will, you will think strange now I should resume, and make good, at least against your objection, that very same proposition. So much of the figure as is needful you will find noted with the same letters, and placed at the end of this fifth lesson. Wherein let B I, be an arch not greater than the radius of the circle, and divided into four equal parts, in L, N, O. Draw S N, the sine of the arch B N, and produce it to T, so as S T be double to S N, that is, equal to the chord B I. Draw likewise _a_ L, the sine of the arch B L, and produce it to _c_, so as _a c_ be quadruple to _a_ L, that is, equal to the two chords B N, N I. Upon the centre N with the radius N I, draw the arch I _d_, cutting B U the tangent in _d_. Then will B N produced cut the arch I _d_, in the midst at _o_. In the line B S produced take S _b_, equal to B S; then draw and produce _b_ N, and it will fall on the point _d_. And B _d_, S T, will be equal; and _d_ T joined and produced will fall upon _o_, the midst of the arch I _d_. Join I T, and produce it to the tangent B U in U. I say, that the straight line I T U shall pass through _c_. For seeing B S, S _b_, are equal, and the angle at S a right angle, the straight lines B N, and _b_ N, are also equal, and the triangles B N _b_, _d_ N _o_ like and equal; and the lines _d_ T, T _o_ equal. Draw _o i_ parallel to _d_ U, cutting I U in _i_; and the triangles _d_ T U, _o_ T _i_ will also be like and equal. Produce S T to the arch _d o_ I in _e_, and produce it further to _f_, so that the line _e f_ be equal to T _e_; and then S _f_ will be equal to _a c_. Therefore _f c_ joined will be parallel to B S.