122 DEMONSTRATIONS of 



But FX, FY, are perpendicular to AX, AY, and it has been 

 fhewn that 2(FB 2 +FC 2 +FD J +FE 2 ) = 4 (FX 2 +FY 2 ). There- 

 fore AX, AY, are the two lines required to be found. 



The conflruclion is obvious, by afluming a point F, which, 

 for the greater fimplicity, may be in one of the given lines, and 

 by defcribing the figure as above. 



* Cor. If from any point parallels be drawn to AB, AC, AD, 

 AE, and to AX, AY, cutting the perpendiculars FB, FC, FD, 

 FE, and FX, FY, in b, c , d, c, and in x, y> 



2(F^-fF^-+F^ 2 +F^) = 4(F^+F/). 



^ LEMMA IV. Fig. IX. 



Let AB, AC, be two Jiraight lines given by pofition, interfering one 

 another in the point A, and from any point D let DB, DC, be drawn 

 perpendicular to AB, AC ; let CB be joined, and bifecled in E, and 

 from E let EF be drawn parallel, and equal to a given flraight line ', 

 through F let GFH be drawn to meet DB and DC, fo as to be 

 bifecled in F, and through G and H let GK, HK, be drawn parallel 

 to AB, AC : the lines GK, HK, will be given by pofilion. 



Through F draw LM parallel to BC, and through B and C 

 draw BL and CM parallel to EF ; join GL, HM ; from A draw 

 AN parallel and equal to EF ; join LN, MN ; through N draw 

 OP parallel to GL ; and join AO, AP. 



Because GF =. FH, and LF =: FM, GL will be equal 

 and parallel to HM ; and becaufe AN is equal and parallel to 

 BL and to CM, the figures AM and AL are parallelograms. 

 Therefore NL is parallel to GK, and NM to HK. Therefore 

 NG and NH are parallelograms, and OG =r NL z= AB ; hence AO 

 is perpendicular to GK ; and, in the fame manner, AP is perpen- 

 dicular 



