126 DEMONSTRATIONS of 



where perpendiculars drawn from it, meet thp lines given by 

 pofition ; and from X let parallels be drawn to thefe lines, 

 meeting the perpendiculars from any point E in f, g, h. 



Since thefe parallels X/; X^, X/6, interfedl one another in 

 the point X, fo as to make all the angles round it equal, they 

 will divide the circumference of the circle which paffes through 

 X and E, into three equal arches fg, gh, hf, (Lemma ^.'). 

 Therefore N, the centre of the circle, is the centre of gravity 

 of the three points f, g, h, and the line YZ, paffing through N, 

 and meeting the circumference, will be a diameter of the cir- 

 cle, and therefore YXZ is a right angle. 



THEOREM XIV. Fig. XII. ^c. 



Lei any number, m, greater than 3, of Jlra'ight lines he given by 

 pofition, three flraight lines may be found, which will he given by 

 poftion, fuch, that if from any point there be drawn perpendiculars 

 to the linet given by poftion, and to the three lines found, thrice the 

 fum of the fqtiares of the perpendiculars to the lines given by pofition, 

 •will be equal to the fum of the fquares of the perpendiculars drawn 

 to three lines found, multiplied by the number m. 



Let to be = 4. 



Cafi I. When the lines (Fig. 12.) AF, BG, CH, DK, 

 given by pofition, are all parallel. Let a perpendicular from any 

 point E meet the parallels in the points A, B, C, D, and let L 

 be the centre of gravity of thefe points. Aflume in AL any 

 point X, and let Y and Z, on the oppofite fide of L, be fuch, that 

 LY-fLZ - LX, and alfo LX^+LY'+LZ' = |(LA^+LB'-|- 

 LC^+LD^) ; then if the aflumed point X be given, the points 

 Y and Z will alfo be given. Draw through the points X, Y, Z, 

 ftraight lines parallel to AF, and they will be the lines required. 



It 



