Mathematical recreations and essays . ll Picture Book, London, 1899, pp. 2G4, 266, where they appear in theform in which I originally gave them. 46 GEOMETRICAL RECREATIONS [CH. Ill 00 — OE. Similarly, since HO bisects OB and DA and is per-pendicular to them, we have OD — OA. Also, by construction,DO = AE. Therefore the three sides of the triangle ODO areequal respectively to the three sides of the triangle , bj Euc. I. 8, the triangles are equal; and therefore theangle ODO is equal to the angle OAE. Again, since HO bisects DA and is perpendicular to it, wehave the angle ODA equal to t
Mathematical recreations and essays . ll Picture Book, London, 1899, pp. 2G4, 266, where they appear in theform in which I originally gave them. 46 GEOMETRICAL RECREATIONS [CH. Ill 00 — OE. Similarly, since HO bisects OB and DA and is per-pendicular to them, we have OD — OA. Also, by construction,DO = AE. Therefore the three sides of the triangle ODO areequal respectively to the three sides of the triangle , bj Euc. I. 8, the triangles are equal; and therefore theangle ODO is equal to the angle OAE. Again, since HO bisects DA and is perpendicular to it, wehave the angle ODA equal to the angle OAD, Hence the angle ADO (which is the difference of ODO andODA) is equal to the angle ^ (which is the difference ofOAE and OAD). But ADO is a right angle, and DAE isnecessarily greater than a right angle. Thus the result isimpossible. Second Fallacy*. To prove that a part of a line is equal tothe whole line. Let ABO be a triangle; and, to fix our ideas,let us suppose that the triangle is scalene, that the angle B is. acute, and that the angle A is greater than the angle 0. FromA draw AD making the angle BAD equal to the angle (7, andcutting BO in D. From A draw AE perpendicular to BO. The triangles ABO, ABD are equiangular; hence, by 19, A ABO : AABD=AO^ : AD\ Also the triangles ABO, ABD are of equal altitude; hence, byEuc. VI. 1, £^ABG : A ABD = BO : BD, .-. AG^.AD^^ AG^ _AD^•• BO BD * See a note by 3T. Coccoz in UIllustration, Paris, Jan. 12, 1895. CH. Ill] GEOMETRICAL RECREATIONS 47 Hence, by Enc. ii. 13, AB^ + BG^- ^AB^-hBB^- 2BD, BEBG Bi) /. ^^BG-^BE^^ + BD- 2BE. AB^ ^AB- . AB^^-BG,BDBG BD .-. BG=^BD, a result which is impossible. Third Fallacy*. To prove that the sum of the lengths of twosides of any triangle is equal to the length of the third side, A,._, , , ,0
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