ii6 GEOMETRICAL PORISMS. 



FCE. Therefore, that the propofition may be univerfally true, 

 tliefe four circles muft interfecSl one another at the fame point. 



About any two of thefe triangles, as ABC, DBE, let circles 

 be defcrlbed, the point P muft be at their interfe<ftion. 



Because ADF is a triangle, and through two of its angles 

 A, D, circles are defcribed, meeting each other at B, a point in 

 AD, therefore (Lemma.) P, tlieir other interfedlion, and the 

 points F, C, E, are in a circle ; and becaufe FCE is a triangle, 

 and circles pafs through C, E, two of its angles, and meet each 

 other at B, a point in CE, therefore (Lemma.) the points P, A, 

 D, F are in a circle. Thus, it appears, that circles defcribed 

 about each of the four triangles ADF, ABC, DBE, CFE, pafs 

 throiTgh the fame point P as was to be inveftigated. It remains 

 to inquire, whether the ratios of GH, HK, KL to one another 

 be given. Join PB, PC, PE, alfo PG, PH, PK, PL. The angle 

 GPH is equal to GAH, that is, to BPC, and PGH is equal to 

 PBC, therefore the triangles BPC, GPH are fimilar, and the an- 

 gle PHK is equal to PCE; but HPK is equal to HFK, that is, to 

 CFE or CPE, hence the triangles HPK, CPE are fimilar, and 

 PKL is equal to PEL. Now, if PN be drawn, fo that the angle 

 BPN may be equal to GPL, that is, to the given angle GBL, it 

 is evident that the point N is given, and will be in a circle paf- 

 fing through P, and touching AG at B ; the angles NPE, LPK 

 will thus be equal, and the triangles NPE, LPK fimilar. Since, 

 therefore, the triangles BPC, CPE, EPN are fimilar to GPH, 

 HPK, KPL, it follows, that BN, CtL are fimilarly divided by the 

 given lines CH, EK, therefore the ratios of GH, HK, KL are 

 the fame witli the given ratios of BC, CE, EN. 



CoNSTnucTTON. About ABC, DBE any two of the four 

 triangles formed by the given lines, let circles be defcribed, they 

 will meet each other at P, tlie point which is to be found. 



THROUGH 



