CH. Ill] 



GEOMETRICAL RECREATIONS 



45 



series: two containing easy theorems or problems, and the 

 third consisting of geometrical fallacies, the errors in which 

 the student was required to find. 



The collection of fallacies prepared by Euclid is lost, and 

 tradition has not preserved any record as to the nature of the 

 erroneous reasoning or conclusions; but, as an illustration of 

 such questions, I append a few demonstrations, leading to 

 obviously impossible results. Perhaps they may amuse any one 

 to whom they are new. I leave the discovery of the errors to 

 the ingenuity of my readers. 



First Fallacy*- To prove that a right angle is equal to an 

 angle which is greater than a right angle. Let ABCD be a 

 rectangle. From A draw a line AE outside the rectangle, 

 equal to AB or DO and making an acute angle with AB, as 



0.— 



indicated in the diagram. Bisect GB in H, and through H 

 draw HO at right angles to GB. Bisect GE in K, and through 

 K draw KO at right angles to GE. Since GB and GE are not 

 parallel the lines HO and KO will meet (say) at 0. Join OA, 

 OE, OG, and OD. 



The triangles OBG and OAE are equal in all respects. 

 For, since KO bisects GE and is perpendicular to it, we have 



» I believe that this and the fourth of these fallacies were first published 

 in this book. They particularly interested Mr C. L. Dodgson; see the Lewis 

 Carroll Picture Book, London, 1899, pp. 264, 266, where they appear in the 

 form in which I originally gave them. 



