60 INVESTIGATION of fome 
is then equal to half the cube on GD, or 2GF —2Gv = 
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, fhall be equal to the 
cube on the given line. 
Ler AC be the given line, (Pl. II. Fig. 6.). With A 
as radius, defcribe an arc AB. Take the angle ACB = 36°. 
Draw AG perpendicular to CB, and join AB. From A and C 
‘ : .. AB ‘ . 
as centres, defcribe arcs with the radii ria and CG, interfecting 
in the point F. Then CFA is the triangle required; and 
2-OF —2-AF =CA- 
DEMONSTRATION. 
Since the angle ACB is 36°, AB is the fide of a decagon in- 
{cribed in the circle, which has AC for its radius ; and CG is the 
perpendiculaf to the fide of an infcribed pentagon. Butitis _ 
AGH AB and AC = AB + 
well known, that CG is = 
ACXAB. Confequently 3AC° = 3AC’x AB + 3ACX AB; 
add AC’ to both, and we have 4 AG’ = AC’ + 3 AG x AB + 
3AC 
