CH. Ill] GEOMETRICAL RECREATIONS 49 



OE perpendicular to AG produced. Draw OF perpendicular 

 to AB produced. Join OB, OG. 



Following the same argument as before, from the equality 

 of the triangles AOF and AOE, we obtain AF= AE; and, 

 from the equality of the triangles BOF and GOE, we obtain 

 FB = EG. Therefore AF-FB = AE- EG, that is, AB - A G. 



Thus in all cases, whether or not DO and AO meet, and 

 whether they meet inside or outside the triangle, we have 

 AB=AG: and therefore every triangle is isosceles, a result 

 which is impossible. 



Fifth Fallacy*. To prove that 7r/4 is equal to v/3. On the 

 hypothenuse, BG, of an isosceles right-angled triangle, DBG, 

 describe an equilateral triangle ABG, the vertex A being on 

 the same side of the base as D is. On GA take a point H so 

 that CH = CD. Bisect BD in K. Join HK and let it cut GB 

 (produced) in L. Join DL. Bisect DL at M, and through 

 M draw MO perpendicular to DL. Bisect HL at N, and 

 through N draw NO perpendicular to HL. Since DL and HL 

 intersect, therefore MO and NO will also intersect ; moreover, 

 since BDG is a right angle, MO and NO both slope away from 

 DG and therefore they will meet on the side of DL remote 

 from A. Join OG, OD, OH, OL. 



The triangles OMD and OML are equal, hence OD = OL. 

 Similarly the triangles ONL and ONH are equal, hence 

 OL = OH. Therefore OD = OH. Now in the triangles OGD 

 and OGH, we have OD = OH, CD = GH (by construction), and 



* This ingenious fallacy is due to Captain Turton : it appeared for the first 

 time in the third edition of this work. 



B. R, i 



