50 



GEOMETRICAL RECREATIONS 



[CH. Ill 



OG common, hence (by Euc. I. 8) the angle OGB is equal to the 

 angle OGH. Hence the angle BCD is equal to the angle BGH, 

 that is, 7J-/4 is equal to tt/3, which is absurd. 



Sixth Fallacy*- To prove that, if two opposite sides of a 

 quadrilateral are equal, the other two sides must be parallel. 

 Let ABGB be a quadrilateral such that AB is equal to BG. 

 Bisect AB in M, and through M draw MO at right angles to 

 AB. Bisect BG in N, and draw NO at right angles to BG. 



If MO and NO are parallel, then AB and BG (which are at 

 right angles to them) are also parallel. 



If MO and NO are not parallel, let them meet in ; then 

 must be either inside the quadrilateral as in the left-hand 



O 



\ 



B N C 



diagram or outside the quadrilateral as in the right-hand 

 diagram. Join OA, OB, OG, OB. 



Since OM bisects AB and is perpendicular to it, we have 

 OA = OB, and the angle 0AM equal to the angle OBM. 

 Similarly OB = OG, and the angle OBN is equal to the angle 

 OGN. Also by hypothesis AB = BG, hence, by Euc. I. 8, the 

 triangles OAB and OBG are equal in all respects, and therefore 

 the angle AOB is equal to the angle BOG. 



Hence in the left-hand diagram the sum of the angles 

 AOM, AOB is equal to the sum of the angles BOM, BOG; 

 and in the right-hand diagram the difference of the angles 

 AOM, AOB is equal to the difference of the angles BOM, BOG; 

 and therefore in both cases the angle MOB is equal to the 

 angle MOG, i.e. OM (or OM produced) bisects the angle BOG. 

 But the angle NOB is equal to the angle NOG, i.e. ON bisects 

 the angle BOG; hence OM and ON coincide in direction. 



* Mathesit, October, 1893, series 2, vol. m, p. 224. 



