INVESTIGATION of PORISMS. 163 



Hence this conftruction : Having drawn HD perpendicular 

 to D E, and D L touching the circle ABC, make D K and 

 D K' each equal toDL, and find G the centre of a circle de- 

 fcribed through the points K, F and K' ; that is, let F K.' be 

 joined, and bifecled at right angles by the line M N, which meets 

 D E in G ; G will be the point required, or it will be fuch a 

 point, that if G B be drawn from it, touching the circle ABC, 

 and G F to the given point, G B and G F will be equal to one 

 another *. 



The fynthetical demonftration needs not be added ; but it 

 is neceflary to remark, that there are cafes in which this con- 

 ftrudtion fails altogether. 



For, firft, if the given point F be any where in the line 

 H D, as at F', it is evident, that M N becomes parallel to D E, 

 and that the point G is no where to be found, or, in other 

 words, is at an infinite diftance from D. 



This is true in general ; but if the given point F coincide 

 with K, then the line M N evidently coincides with D E ; fo 

 that, agreeably to a remark already made, every point of the 

 line D E may be taken for G, and will fatisfy the conditions of 

 the problem ; that is to fay, G B will be equal to G K, wherever 

 the point G be taken in the line D E. The fame is true if F 

 coincide with K'. 



This is eafily demonstrated fynthetically ; for if G be any 

 point whatfoever in the line D E, from which G B is drawn 

 touching the circle ABC; if D K and D K' be each made e- 

 qual to D L ; and if a circle be defcribed through the points 

 B, K, and K' ; then, fince the rectangle K H K', together with 

 the fquare of D K, that is, of D L, is equal to the fquare of 



X 2 DH, 



* This folution of the problem was fuggefted to me by Profeflbr Robison ; and is 

 more fimple than that which I had origiiplly given. 



