48 GEOMETRICAL RECREATIONS [CH. Ill 



get smaller and smaller. Finally the points G, J, L,... will 

 approach indefinitely near the line A 0, and ultimately will lie 

 on it; when this is the case the sum of AG and GH will be 

 equal to AM, and similarly for the other similar pairs of lines. 

 Thus, ultimately, 



AB + BG = AH + HK + KM + MG 



= A0, 



a result which is impossible. 



Fourth Fallacy. To prove that every triangle is isosceles. 

 Let ABO be any triangle. Bisect BO in B, and through B 

 draw BO perpendicular to BO. Bisect the angle BAG by AO. 



First. If BO and AO do not meet, then they are parallel. 

 Therefore AO is at right angles to BO. Therefore AB = AC. 



Second. If BO and AO meet, let them meet in 0. Draw 

 OE perpendicular to AG. Draw OF 

 perpendicular to AB. Join OB, 00. 



Let us begin by taking the case 

 where is inside the triangle, in 

 which case E falls on AG and F on 

 BG. 



The triangles AOF and AOE are 

 equal, since the side AO is common, 



angle OAF= angle OAE, and angle OF A = angle OEA. Hence 

 AF=AE. Also, the triangles BOF and GOE are equal. For 

 since OB bisects BG at right angles, we have OB = OG; also, 

 since the triangles AOF and AOE are equal, we have 

 0F= OE; lastly, the angles at F and E are right angles. 

 Therefore, by Euc. I. 47 and I. 8, the triangles BOF and GOE 

 are equal. Hence FB = EG. 



Therefore AF+FB = AE + EG, that is, AB = AG. 



The same demonstration will cover the case where BO and 

 AO meet at B, as also the case where they meet outside BG 

 but so near it that E and F fall on AG and AB and not on 

 AG and AB produced. 



Next take the case where BO and AO meet outside the 

 triangle, and E and F fall on iO and AB produced. Draw 



