46 



GEOMETRICAL RECREATIONS 



[CH. Ill 



OG = OE. Similarly, since HO bisects GB and DA and is per- 

 pendicular to them, we have OD = OA. Also, by construction, 

 DC = AE. Therefore the three sides of the triangle ODG are 

 equal respectively to the three sides of the triangle OA E. 

 Hence, by Euc. I. 8, the triangles are equal; and therefore the 

 angle ODG is equal to the angle OAE. 



Again, since HO bisects DA and is perpendicular to it, we 

 have the angle OD A equal to the angle OAD. 



Hence the angle ADG (which is the difference of ODG and 

 ODA) is equal to the angle DAE (which is the difference of 

 OAE and OAD). But ADG is a right angle, and DAE is 

 necessarily greater than a right angle. Thus the result is 

 impossible. 



Second Fallacy*- To prove that a part of a line is equal to 

 the whole line. Let ABG 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 G. From 

 A draw AD making the angle BAD equal to the angle C, and 

 cutting BG in D. From A draw AE perpendicular to BG. 



The triangles ABG, ABD are equiangular; hence, by Euc. 

 vi. 19, A ABG : A ABD =AG' : AD\ 



Also the triangles ABG, ABD are of equal altitude ; hence, by 

 Euc. vi. 1, 



A ABG : AABD=BG : BD, 



.-. AC>:AD*=BC : BD. 



. AG * AD ' 

 " BG ~ BD' 



* See a note by M. Cocooz in V Illustration, Paris, Jan. 12, 1895. 



