ii8 GEOMETRIC JL PORISMS.' 



of any two of the lines, a circle be defcribed to meet them 

 in K and L, a line joining KL, and meeting the remaining 

 lines, will be divided by them into fegments HK, KL, LM, 

 MN, eft', having to each other given ratios. 



Let Q^ R, S, life, be the remaining interfedlions of GB, FC, 

 ED, ^V. Becanfe GRE is a triangle, and circles PGAB, PiiAD 

 pafs through G, E, two of its angles, and meet at A, a point in 

 GE, the points P, R, B, D, are in a circle, (Lemma.) in the 

 fame way it appears, that circles may pafs through P, S, C, D, 

 and P, Qi^ B, C, i^c. Becaufe it is now proved, that in the tri- 

 angle CDS, a circle may pafs through P, C, B, Q^ and another 

 through P, D, B, R ; therefore the points P, S, R, Q^ are in a 

 circle. (Lemma.) Thus it may be Ihewn, that circles defcribed 

 about each of the triangles, formed by the intercepted fegments 

 of the flraight lines, will all pafs through the fame point P. 

 From P draw flraight lines to the points of interfe<ftion of one 

 of the given lines, with all the others, as PA, PB, PC, PD, ^r. 

 Join PH, PK, PL, PM, PN, ^r. 



Since P, Q^ K, L, are in a circle, the angle BKP is equal to 

 CLP ; now, the angles PBG, PCF, are each equal to PAG ; 

 therefore the angles PBK, PCL, are equal, and the triangles 

 PBK, PCL, fimilar ; hence KP is to PL as BP to PC ; now the 

 angle KPL is equal to KQL, that is, to BPC ; therefore the tri- 

 angles KPL, BPC, are fimilar, and the angle PLM will be equal 

 to PCD. But the points P, S, C, D, having been proved to lie 

 in a circle, if PS be joined, the angle PCD will be equal to PSD, 

 therefore PLM is equal to PSD or PSM, hence the points P, S, 

 L, M are in a circle. In the fame way it may be fhewn, that 

 P, G, H, K are in a circle, as alfo P, D, M, N, ^c. and that the 

 triangles PAH, PDM, Ij^c, are each fimilar to PBK and PCL, 

 and hence that PHK is fimilar to PAB,'and PLM to PCD, l^c 

 Through P defcribe a circle to touch AG at A, and meet AD 



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