120 On tJie DiipUcat'ion of the Cube. 



ed its solution by right lines, or merely by the principles 

 of common geometry. His construction is as follows. 



* " Let ACE, ECD, and DCB be equal angles 

 of any magnitude, and let AC and BC be the ex- 

 tremes given. Draw AB a right line crossing EC and 

 DC in tlie points E and D ; then will AEC=EDC-f 

 DCE, and for the same reason CBD=CDE— DCE. 

 Wherefore make BDF=DCE and AEG=DCE, and 

 we shall have three similar triangles FCD, DCE and 

 ECG, and their sides are necessai'ily proportional ; and 

 the lines CF, CD, CE and CG form a series of four con- 

 tinued proportionals ; for CD is the hypothenuse of the 

 first triangle, and the base of the second ; and is there- 

 fore a mean between CF and CE. In like manner CE 

 is a mean betvv^een CD and CG. But the extremes CF 

 and CG are shorter than CB and CA. Having there- 

 fore by this process ascertained the method of finding 

 easily four continued proportionals by beginning with 

 the mean terms ; if we invert the process, and begin 

 with the extremes, and make the angles EAI and DBK 

 each equal to DCE, we shall have BK parallel to FD, 

 and AI parallel to EG, and therefore KI parallel to DE. 

 Therefore, the triangles CBK, CKI, and CIA are simi- 

 lar to CDE ; and by reason of position, CK and CI are 

 the mean proportionals sought." Thus far it appeared 

 necessary to transcribe Mr. Winthrop's solution of this 

 problem. The demonstration appears faultless until we 

 come to the words, and therefore KI parallel to DE. 

 From vvhat premises this inference is drawn, is not 

 easily conceived. It certainly cannot be justified by 

 any principles antecedently expressed ; for the parallel- 

 ism of BK to FD, and of AI to EG contains no relation, 

 whereby the parallelism of KI to DE may be inferred, 

 and unless some other principles were taken, into con- 

 sideration, it is evident, that the author has fallen into a 

 paralogism. 



But that KI is not parallel to DE, except when the 

 antecedents and consequents of the proportionals, or the 

 two extremes are equal, may be easily proved ; for 

 which purpose make the angle EGH=FDB, and from 

 H draw HK, then will HK be parallel to ED, and th« 



* See Fig. 1, plate 1. 



