6"o INVESTIGATION of fomc 



is then equal to half the cube on GD, or 2 GF — 2 GV = 



gd\ 



Hence an eafy folution of this problem. 



Having two equal right lines given, it is required to cut one 

 of them into two parts, and the other into three parts ; fo that 

 the cubes on the two parts, into which the one of thefe lines is 

 cut, fhall, together, be equal to the cubes on the three parts, in- 

 to which the Other is cut, taken together. 



Hence, alfo, an eafy conftruction for this problem: On a gi- 

 ven right line, to conftitute a triangle, fuch that twice the dif- 

 ference of the cubes on the other two fides, lhall be equal to the 

 cube on the given line. 



Let AC be the given line, (PI. III. Fig. 6.). With A 

 as radius, defcribe an arc AB. Take the angle ACB — ^6°» 

 Draw AG perpendicular to CB, and join AB. From A and G 



AB 

 as centres, defcribe arcs with the radii — , and CG, interfering 



2 



in the point F. Then CFA is the triangle required ; and 

 2 • CF 3 — 2 . AF' = CA * 



Demonstration. 



Since the angle ACB is $6°, AB is the fide of a decagon in- 

 fcribed in the circle, which has AC for its radius ; and CG is the 

 perpendiculaf to the fide of an infcribed pentagon. But it is 



AC + AB * z 



well known, that CG is = , and AC = AB + 



AC X AB. Confequently 3AC 3 = 3 AC 2 X AB -j- 3 AC X A3; 



add AG 3 to both, and we have 4 AC 3 = AC 3 -f- 3 AC X AB -j- 



3 AC 



