158 



Selectiom from Foreign Science. 



elementary, yet the frequent occasion to divide a right line, and 

 above all, the singularity of the result at which he had arrived, 

 seemed sufficient to excite the curiosity of the lovers of raathe- 

 niaiics. 



Problem. To divide a right line into any number of equal 

 parts. Let AB. fig. 1. be the line given to be divided into equal 

 parts : 



from its extremity A, extend at any angle the indefinite line Ac, 

 on which are to be laid down the equal parts Aa, ah, be, &c. of an 

 arbitrary length : let ?« be the number of parts from A to D ; join 

 the points D and B by the line DB. prolonged towards F, so that 

 BF=:BDx« (» being any number whatever, whole or fractional.) 

 Then join the point F with the extremity E of any one of the parts 

 on AC, and suppose that AE contains Zparts, then ED will contain 

 (?« — /) of these parts. 



To ascertain in what manner the line AB is divided at G by EF, 

 draw EH parallel to DF, which will give BH : AB :: DE : AD :: 



(m-l) : OT;from which BH=ii5i!^zil. But because of the similar 



m 



triangle AEH, ADB, EH : BD :: AE : AD :: Z : m, and from which 

 EH =: '- ; but as by the construction BF = BD X n, and con- 



sequently BD = 



equal Hi/. In 

 }iin 



BGF, EGH, BF 

 BF. Z 



BF 



by substituting this value for BD, EH will 



BF 



consequence of the similarity of the triangles 



EH : : BG : HG, or, which is the same thing, 

 HG. Dividinsfthe two terms of the first ratio 



BG 



rnn 



by FB, and then dividing them by nm, the proportion will become 

 BG : HG :: mw : I, from which it follows that (BG + HG) : BG 

 ::[mn + I) : mn, or BH : BG ;; [mn + I] : mn. Substituting in 



