114 GEOMETRICAL P R 1 S M S. 



alfo ill a circle ;. therefore the angle HCF is equal to HEF, that 

 is, (as has been fhewn), to R A15 ; hence the point H, which may 

 be found, is in a circle pafling through the points A, B, C, what- 

 ever be the given ratio of DE to EF. Let this circle be defcribed. 



Because the points H, A, D, E are in a circle, the angle 

 HAC is equal to EIDE, and becaufe H, C, E, F are in a circle, 

 the angle HFE is equal to HCA ; therefore the triangles AHC, 

 DHF are fimilar. In the fame manner it appears, that AHB is 

 fimilar to EHF, and CHB to EHD. 



Let AC be divided at K, fo that AK may be to KC, in the 

 given ratio of DE to EF, the point K will thus be given. Join 

 HK meeting the circle in G. The triangles AHC, DHF being 

 fimilar, and -having AC, DF, fimilarly divided at K, E, the tri- 

 angles AHK, KHC will therefore be fimilar to DHE, EHF, 

 which have been proved fimilar toBHC, AHB; therefore the 

 angle AHB is equal to CHK or CHG, and the arch AB is equal 

 to CO, hence G is a given point, and K being given, the fine 

 GH will be given by pofition ; therefore the point H is given 

 which was to be found. 



Construction. Defcribe a circle through the points A, B, C, 

 let AB, BC, be the lines upon which D and F, the extremities of 

 the indeterminate line, are to be placed, and let AC be the line 

 which is to meet it in E, fo that DE may be to EF, in the given 

 ratio of de to ef. Find K, fo that AK may be to KC as ^f to 

 ef, draw BG parallel to AC, meeting the circle in G, join GK 

 meeting the circle in H, the point which may be found ; that is, 

 if any circle be defcribed through H, and B the interfe<ftion of 

 any two of the given lines, to meet them in D and F, and if 

 DF be joined, meeting the remaining line at E, the line DF 

 fhall be divided at E, fimilarly to the given line def. 



Let AH, BH, CH be joined, alfo DH, EH, FH. The angle 

 HDF or HDE is equal to ElBF, that is, to HAE, the points 

 H, A, D, E are therefore in a circl,e, now the points H, B, D, F 



are. 



