“INVESTIGATION of PORISMS. 165 
maining perpendicular ; or, that A Eand BG together, may 
be equal to CF, : 
Suppose it done: Bifect AB in H, join CH, and draw HE 
perpendicular to DG. 
Because AB is bifected in H, the two perpendiculars A E 
_ and B G are together double of H K ; and as they are alfo equal 
to CF by hypothefis, CF muft be double of HK, and CL of. 
LH. Now, CH 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. _ 
_ Tue conftruétion is obvious. Bife@ AB in H, join CH, 
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. Turis 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 point 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 
