INVESTIGATION of PORISMS. 165 



maining perpendicular ; or, that A E and B G together, may- 

 be equal to C F. 



Suppose it done : Bifecl A B in H, join C H, and draw HK 

 perpendicular to D G. 



Because AB is bifected in H, the two perpendiculars AE 

 and B G are together double of H K ; and as they are alfo equal 

 to C F by hypothecs, C F muft be double of HK, and C L of. 

 L H. Now, C H is given in pofition and magnitude ; there- 

 fore the point L is given ; and the point D being alfo given, 

 the line D L is given in pofition, which was to be found. 



The conftrudtion is obvious. Bifedt A B in H, join C H, 

 and take H L equal to one- third of CH ; the ftraight line which, 

 joins the points D and L is the line required. 



Now, it is plain, that while the triangle ABC remains the. 

 fame, the point L alfo remains the fame, wherever the point D 

 may be. The point D may therefore coincide with L ; and 

 when this happens, the pofition of the line to be drawn is left 

 undetermined ; that is to fay, any line whatever drawn through 

 L will fatisfy the conditions of the problem. 



Here therefore we have another indefinite cafe of a problem, 

 and of confequence another Porifm, which may be thus enun- 

 ciated : " A triangle being given in pofition, a point in it may be 

 found, fuch, that any ftraight line whatever being drawn 

 through that point, the perpendiculars drawn to this ftraight 

 line from the two angles of the triangle which are on one fide 

 of it, will be together equal to the perpendicular that is drawn 

 to the fame line from the angle on the other fide of it." 



11. This Porifm may be made much more general; for if, 

 inftead of the angles of a triangle, we fuppofe ever fo many 

 points to be given in a plane, a po'int may be found, fuch, that 

 any ftraight line being drawn through it, the fum of all the 

 perpendiculars that fall on that line from the given points on 



one 



