INVESTIGATION of PORISMS. 163 
Hence this conftruction: Having drawn H D perpendicular 
to DE, and DL touching the circle ABC, make DK and 
DK’ each equal to DL, and find G the centre of a circle de- 
fcribed through the points K, F and K’; that is, let FK’ be 
joined, and bifected at right angles by the line M N, which meets 
DEinG; Gwwill be the point required, or it will be fuch a 
point, that if GB be drawn from it, touching the circle ABC, 
and GF to the given point, G B and G F will be equal to one 
another *. ; 
Tue fynthetical demonftration needs not be added; but it 
is neceflary to remark, that there are cafes in which this con- 
ftruction fails altogether. 
For, firft, if the given point F be any where in the line 
HD, as at F’, it is evident, that MN becomes parallel to DE, 
and that the point G is no where to be found, or, in other 
words, is at an infinite diftance from D. 
Tuils is true in general; but if the given point F coincide 
with K, then the line MN evidently coincides with DE; fo 
that, agreeably to a remark already made, every point of the 
line DE may be taken for G, and will fatisfy the conditions of 
the problem ; that is to fay, G B'will be equal to GK, wherever 
the point G be taken inthe line DE. The fame is true if F 
coincide with K’. 
Tuis is eafily demonftrated fynthetically ; for if G be any 
_ point whatfoever in the line DE, from which GB is drawn 
touching the circle ABC; if DK and DK’ be each made e- 
- qualto DL; andif a circle be defcribed through the points 
|B, K, and K’; then, fince the rectangle K HK’, together with 
B the fquare of DK, that i is, of DL, is equal to the fquare of 
q X 2 DH, 
=! iy 
* This folution of the problem was fuggefted to me by Profeflor Rosison ; and is 
___ more fimple than that which I had originally given. 
