Mathematical recreations and essays . ions in geometryassociated with the name of Euclid, but it is not known sogenerally that these propositions were supplemented originallyby certain exercises. Of such exercises Euclid issued three CH. Ill] GEOMEmiCAL RECREATIONS 45 series: two containing easy theorems or problems, and thethird consisting of geometrical fallacies, the errors in whichthe student was required to find. The collection of fallacies prepared by Euclid is lost, andtradition has not preserved any record as to the nature of theerroneous reasoning or conclusions; but, as an illustrati
Mathematical recreations and essays . ions in geometryassociated with the name of Euclid, but it is not known sogenerally that these propositions were supplemented originallyby certain exercises. Of such exercises Euclid issued three CH. Ill] GEOMEmiCAL RECREATIONS 45 series: two containing easy theorems or problems, and thethird consisting of geometrical fallacies, the errors in whichthe student was required to find. The collection of fallacies prepared by Euclid is lost, andtradition has not preserved any record as to the nature of theerroneous reasoning or conclusions; but, as an illustration ofsuch questions, I append a few demonstrations, leading toobviously impossible results. Perhaps they may amuse any oneto whom they are new. I leave the discovery of the errors tothe ingenuity of my readers. First Fallacy*, To prove that a right angle is equal to anangle which is greater than a right angle. Let A BCD be arectangle. From A draw a line AE outside the rectangle,equal to -45 or DC and making an acute angle with AB, as. indicated in the diagram. Bisect CB in //, and through Hdraw HO at right angles to GB. Bisect CE in K, and throughK draw KO at right angles to CE. Since CB and CE are notparallel the lines HO and KO will meet (say) at 0. Join OA,OE, OC, and OD. The triangles ODG and DAE are equal in all , since KO bisects CE and is perpendicular to it, we have * I believe that this and the fourth of these fallacies were first publishedin this book. They particularly interested Mr C. L. Dodgson; see the LewisCarroll Picture Book, London, 1899, pp. 2G4, 266, where they appear in theform in which I originally gave them. 46 GEOMETRICAL RECREATIONS [CH. Ill 00 — OE. Similarly, since HO bisects OB and DA and is per-pendicular to them, we have OD — OA. Also, by construction,DO = AE. Therefore the three sides of the triangle ODO areequal respectively to the three sides of the triangle , bj Euc. I. 8, the triangles are equal; and therefore theangle
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