Mathematical recreations and essays . CH. Ill] GEOMETRICAL RECREATIONS 49 OE perpendicular to AG produced. Draw OF perpendicularto AB produced. Join OB, 00. Following the same argument as before, from the equalityof the triangles A OF and AOE, we obtain AF=AE\ and,from the equality of the triangles BOF and GOE, we obtainFB = EG. Therefore AF-FB=AE-EG, that i&^AB^ Thus in all cases, whether or not BO and AO meet, andwhether they meet inside or outside the triangle, we haveAB—AG\ and therefore every triangle is isosceles, a resultwhich is impossible. Fifth Fallacy^, To prove that irj^ is equ
Mathematical recreations and essays . CH. Ill] GEOMETRICAL RECREATIONS 49 OE perpendicular to AG produced. Draw OF perpendicularto AB produced. Join OB, 00. Following the same argument as before, from the equalityof the triangles A OF and AOE, we obtain AF=AE\ and,from the equality of the triangles BOF and GOE, we obtainFB = EG. Therefore AF-FB=AE-EG, that i&^AB^ Thus in all cases, whether or not BO and AO meet, andwhether they meet inside or outside the triangle, we haveAB—AG\ and therefore every triangle is isosceles, a resultwhich is impossible. Fifth Fallacy^, To prove that irj^ is equal to tt/S. On thehypothenuse, BG, of an isosceles right-angled triangle, DBG,describe an equilateral triangle ABG, the vertex A being onthe same side of the base as D is. On GA take a point H sothat GH = GD. Bisect BD in K. Join HK and let it cut GB(produced) in L. Join DL. Bisect DL at M, and throughM draw MO perpendicular to DL. Bisect HL at N, andthrough N draw NO perpendicular to HL. Since DL and HLintersect, therefore MO and NO will also intersect; moreover,since BDG is a right angle, MO and NO both slope away fromDO and therefore they will meet on the side of DL remotefrom A. Join OG, OD, OH, OL. The triangles OMD and OML are equal, hence OD = OL,Similarly the triangles ONL and ONH are equal, henceOL = OH. Therefore OD = OH. Now in the triangles OGDa
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