Mathematical recreations and essays . 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) /. ^^
Mathematical recreations and essays . 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. Let ABG be a triangle. Complete the parallelogram ofwhich AB and BG are sides. Divide AB into n + 1 equalparts, and through the points so determined draw n linesparallel to BG. Similarly, divide BG into n + 1 equal parts,and through the points so determined draw n lines parallel toAB. The parallelogram ABGD is thus divided into (n+iyequal and similar parallelograms. I draw the figure for the case in which n is equal to 3,then, taking the parallelograms of which AG is a, diagonal, asindicated in the diagram, we have AB + BG = AG + HJ + KL-hMN + GH + JK + LM-\-NG. A similar relation is true however large n may be. Nowlet n increase indefinitely. Then the lines AG, GH, &c. will * The Canterbury Puzzles, by H. E. Dudeney, London, 1907, pp. 26—23. 48 GEOMETRICAL RECREATIONS [CH. Ill get smaller and smaller. Finally the points 0, «/, Z,... willapproach indefinitely near the line AG, and ultimately will lieon it; when this is the case the sum of AO and GH will beequal to AH^ and simil
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