of Polygons. 1 QZ 



LMthe Z A IO = ALM = AHG, therefore ZOIG=EGH = 

 EBD, and a circle may consequently be described about the 

 quadrilateral BIOE; the Z BIE is therefore =:B0P1 In 

 like manner since LP is parallel to IE, the ZMLP = OIE = 

 OBE=DCE = (by const.) ANM, and a circle may be de- 

 scribed about the quadrilateral LNMP; the ZNLM is there- 

 fore = NPM. But I O and I E being parallel to LM and 

 LP, AL : AI : : AM : AO : : AP : AE, therefore OE is pa- 

 rallel to MP and the ZMPA = OEA. The ZBAE which is 

 = AOE + AEO, is therefore =BIE + CHG, and BIA + 

 CHA being thus =BAE+ EGH = BAE-f- DCA= BDC, 

 their supplemental Zl RH will be equal to the angle contained 

 in the segment BFC, and the point of intersection R will 

 therefore be in the circumference of the circle BGCF. q.e.d, 



?=»-n: 



The demonstration now given is equally aj)})licable to any 

 position of the point G in the circumference, and as the line 

 drawn through A is also unlimited in position, the point of 

 intersection R may be either in the same segment with G, or in 

 the opposite, or in the adjacent segments. Wherever, therefore, 

 the two })oints G and R be assumed, if GD GE and RB RC 

 be joined and the opposite lines produced to meet, that is, RB 

 with EG and RC with DG, the points of intersection and the 

 point A will be always in the same straight line. 



In like manner if 111) RE and GB GC be drawn, another 

 set of intersections will be obtained having the point A com- 

 mon to both ; from any two points, therefore, two sets of in- 

 tersections may always be had with reference to the same se- 

 cants; but G and R may also be referred to the secants DE 

 and BC, and tlius two otiier sets be produced in which the in- 

 tersection of the secants is common to both. 



X 2 Accord- 



