ii6 GEOMETRICAL PORISMS. 



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

 thefe four circles muft interfed one another at the fame point. 



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

 be defcribed, the point P mnft be at their interfediion. 



Because ADF is a triangle, and through r«^o of its angles 

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

 AD, therefore (Lemma.) P, their other interfecflion, 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 

 through 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 PflK is equal to PCE; but HPK is equal to HFK, that is, ta 

 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- 

 fmg 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, GL are fimilarly divided by the 

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

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



Construction. About ABC, DBE any two of the four 

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

 w,ill meet each other at P, the point which is to be found. 



THROUGH 



