[II. Prop. 1. Cor. Now each of the angles , , is less than . Lines , , will fall within angles , ; i.e. Figure will fall within Triangle . Join , DC′. Now 'amounts' of Triangles , , together = , and those of the other 4 Triangles are, by our First Hypothesis, together less than ; 'amounts' of all 6 Triangles are together less than ; but these make up 'amount' of Triangle , plus angles at , , , which together = ; 'amount' of Triangle , plus , is less than . Now we know that is not-greater than ; , adding these inequalities, 'amount' of Triangle , plus , is less than ; this 'amount,' alone, is less than ; which is absurd, since the latter lies below the 'inferior limit,' and is therefore an 'impossible amount'; one of our two Hypotheses must be false; i.e. either there is an 'amount' not-less than , or else any 'amount' is not-less than . Suppose we maintain our First Hypothesis: then we must abandon our Second; i.e. we must admit that any 'amount' is not-less than . It shall be proved that, in this case, we must also admit that any 'amount' is not-less than . For, if we deny this, we must assert that there is an 'amount' less than . Let this be our Third Hypothesis. Then, in the above proof, and may be replaced by and , and a similar absurdity will follow. Hence either our First or our Third Hypothesis must be false; i.e. either there is an 'amount' not-less than , or else any amount is not-less than . A similar proof will hold for ; and then for . Hence, either there is an 'amount' not-less than , or else any 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. Corollary. [Pg 21] The 'possible region' does not lie wholly below two right angles. PROP. IV. Theorem. [Pg 22] There is a Triangle whose angles are together equal to two right angles. For the 'possible region' does not lie wholly above ; [I. Prop. 6. Cor. neither does it lie wholly below ; [II. Prop. 3. Cor. it includes . [I. 4. Cor. 2. That is, there is a Triangle &c. Q.E.D. PROP. V. Theorem. There is a quadrilateral Figure which is 'rectangular,' that is, which has all its angles right angles. Let be a Triangle whose 'amount' = . [II. Prop. 4. At make angle equal to angle , and angle equal to angle ; hence angles , , , together = ; , , are in a straight Line. [Euc. I. 14. Bisect at ; from draw perpendicular to ; from cut off equal to ; and join . , in Triangles , , , , are respectively equal to , , and angle to angle , [Euc. I. 4. the Triangles are equal in all respects; angle is a right angle; and angle ; angles , , together = angles , ; i. e. together = ; , are in a straight line; [Euc. I. 14. is a common perpendicular to Lines , . Similarly, by bisecting at , it may be proved that is a common perpendicular. Hence Figure is rectangular.