162 Mr. T. Drummond on a Property 



As this circumstance appeared remarkable, I calculated 

 the position of the planet for the above given time, viz. 

 22^ 53' 29", 5 on June 15, taking the arguments of the tables 

 for 10'^ 53' 29", 5 on June 16, as the epoch of them is mid- 

 night, and found the result very nearly corresponding to that 

 mentioned by Professor Encke ; the small difference having 

 been occasioned probably by having used a different ephe- 

 meris for the sun's longitude, &c. I shoidd esteem it a favour 

 if any gentleman would explain how the tables could give the 

 longitude at the above opposition within so small an erroi-, if 

 they are twenty minutes in arrear. 



March 8, 1823. W. M. M. 



XXXV. On a Properti) of Polygons. By T. Drummond, Esq. 

 To the Editors of the Philosophical Magazitie a?id Jovrnal. 



npiIE following proposition, together with the more general 

 -*■ })roperty deduced from it, not being given, I believe, in 

 any of our books on elementary geometry, a demonstration 

 may, perhaps, not be uninteresting to some of your mathema- 

 tical readers. I am 



Your most obedient servant. 

 Tower, Feb. 14, 1823. T. Drummond. 



If the opposite sides of an irregular hexagon inscribed in a 

 circle be produced to meet, the three points of intersection 

 will be in the same straight line. 



The demonstration which follows, is of the converse of this 

 proposition, from which the one now enunciated may be de- 

 duced as a corollary. 



Let a circle and a point A without it be given ; if any two 

 secants AB and AC be drawn, and straight lines EG and DG 

 be also drawn through any point G of the intercepted arc to 

 meet when produced any straight line drawn through A, then 

 if the points of intersection IB and HC be joijied, and IB 

 and HC produced to meet, their intersection R will be in the 

 circumference of the circle BDGEC. 



Produce AB ind AC till AM and AN be respectively 

 equal to AD and AC, make AL = AH, and join MN, ML 

 and NL; the lines ML and NL will thus be parallel to DH 

 and CH, and the angles ML A and NLA= respectively to 

 AHD and AHC. From L draw LP parallel to IE, and 

 meeting AC in P, from 1 draw lO parallel to LM, and meet- 

 ing AM in O, join EC) and MP. Since 10 is parallel to 



LM 



