Afis'wer to Mr. Harnett's Question. 309 



which is mucli at your service, if it should be thought to merit 

 insertion in your vahiable Journal. 



I am your obedient servant, 



Paul Newton. 



In prop. 47, produce DB to meet FC in M, and EC to 

 meet KB in N. From the point E draw EO parallel to 

 CM; and from D draw DP parallel to BN (prop. 31st): 

 Then will the parallelograms DBCE, DBNP, and ECOM, 

 be equal to one another 

 (props. 35 and 36). The 

 ])arallelograms D B T L, 

 DBRS, and OMRS, are^ 

 also equal to one another. 

 For the same reason, the 

 parallelograms L T C E, 

 SRCE, and SRNP, are 

 equal to one another. Now 

 LT is equal to SR (prop. 

 34) ; consequently L S is 

 equal to T R. But the side 

 SR is evidently common 

 to the four parallelograms 

 DBRS, OMRS, ECRS, 

 and PNRS; therefore the 

 lines FC, KB, and AL, intersect one another in the point 

 R; or otherwise the opposite sides of parallelograms could 

 not be equal. But further; draw BL, BS, CL, and CS. 

 The triangle FBC is equal to the triangle BDL (props. 34 

 and 47), or equal to the triangle BSR (props. 37 and 38). 

 For the same reason, the triangle BKC is equal to the trian- 

 gle CLE, or is equal to the triagle CSR. The three triangles 

 LDS, LBS, and TBR, are equal to one another (props. 37 

 and 38). Again, the three triangles LES, LCS, and TCR, 

 are equal to one another. Hence we perceive that the two tri- 

 angles LBS and LCS, which meet in the point S, on the line 

 LA, are respectively equal to the two triangles DLS and ELS, 

 which meet likewise in the point S ; or they are respectively 

 equal to the two triangles TBR and TCR, whose sides B R 

 and C R meet in the point R, on LA. Besides, we perceive 

 that because OM is ecjual to BD, BM is equal to OD. 

 Therefore the triangles MRB and O S D are equal to each 

 other, and the triangles N R C and P S E are for a similar 

 reason ecjual to each other (prop. 38); consequently the lines 

 FC, KB, and A L, intersect one another in the point R. 



o. E. D. 

 SOLUTiON 



