Note.—The Reader is requested to imagine a chord drawn to the arc . [Pg 37] Hence it may be proved, as in Book II, Prop. I, that any Chord, drawn from to any Point on the Arc , is greater than , and therefore not-less than . Now, on our Hypothesis, is a Sector whose Chord is not-less than its Radius; , in a Sector whose vertical angle is , its outer Segment is greater than its central Triangle; [Prop. A. i.e., in a Sector whose vertical angle is , each of the equal sides of its inscribed isosceles Triangle is not-less than its Radius, and its outer Segment is greater than once its central Triangle; , in a Sector, whose vertical angle is , its outer Segment is greater than twice its central Triangle; [Prop. B. , similarly, in a Sector, whose vertical angle is , its outer Segment is greater than 4 times its central Triangle; and so on; , ultimately, in a Sector, whose vertical angle is , its outer Segment is greater than times its central Triangle; but ; , in Sector , Segment is greater than times Triangle . But this is absurd, since it has been already proved less than times this Triangle. Hence our Hypothesis, that is not-less than , is false; i. e. is less than . Therefore an isosceles Triangle &c. Q.E.D. Corollary to Prop. C. [Pg 38] Hence, by Book I, Prop. III, it is possible to describe, on a given base, an isosceles Triangle having each base-angle equal to of a right angle. [N.B. If the Reader be willing to grant, as axiomatic, that 1024 times the Tetragon is greater than the Segment, he must now admit, as logically proved, that an isosceles Triangle, whose vertical angle is of a right angle, has its base less than either of its sides. If such a Triangle were actually drawn, having each side 140 yards long, its base would be found to be less than an inch!] PROP. D. Theorem. [To be substituted for Prop. II, at p. 18.] The angles of any Triangle are together not-less than of a right angle. [The proof of Prop. II will serve here, without any change, except the substitution of '' for ''.] [N.B. If the Reader be willing to grant, as axiomatic, that 1024 times the Tetragon is greater than the Segment, he must now admit, as logically proved, that the angles of any Triangle are together not-less than of a right angle.] PROP. E. Theorem. [Pg 39] [To be substituted for Prop. III, at p. 19.] There is a Triangle whose angles are together not-less than two right angles. [The proof of Prop. III will serve here, without any change, except the substitution of '' for ',' and '' for 'one-eighth,' down to the words 'A similar proof &c.' at foot of p. 21; for which the following is to be substituted.] A similar proof will hold for , , and so on; and ultimately for . Hence, either there is an 'amount' not-less than , or else every 'amount' is not-less than . But . Hence the second clause of this alternative contains the first. Hence the first clause must be true. That is, there is a Triangle &c. Q.E.D.