CH. Ill] GEOMETRICAL RECREATIONS 47 



Hence, by Euc. n. 13, 



AB* + BC* -2BG.BE AB' + BD' - 2BD.BE 

 BG ~ BJD ' 



y)Di A Ra 



.'.j^ + BC-2BE = ^L + BD-2BE. 

 . AB' JtT) _AB» B0 



. AB' -BG.BD _ AB* -BC.BD 

 BG BD 



.-. BG = BD, 



a result which is impossible. 



Third Fallacy*. To prove that the sum of the lengths of two 

 sides of any triangle is equal to the length of the third side. 

 A K , , , 7 d 



Let ABC be a triangle. Complete the parallelogram of 

 which AB and BG are sides. Divide AB into n + 1 equal 

 parts, and through the points so determined draw n lines 

 parallel to BG. Similarly, divide BG into n + 1 equal parts, 

 and through the points so determined draw n lines parallel to 

 AB. The parallelogram ABGD is thus divided into (n + 1)' 

 equal 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, as 

 indicated in the diagram, we have 



AB + BG = AG + EJ + KL + MN 

 + GH + JK + LM + NO. 

 A similar relation is true however large n may be. Now 

 let n increase indefinitely. Then the lines AG, GH, &c. will 

 * The Canterbury Puzzles, by H. E. Dudeney, London, 1907, pp. 26—23. 



