jMifcellanea Curio fa. 6 1 



But here it muft be noted, that if the fum 

 of the obferved angles, B A E, A E £>, is i$o 

 degrees : then A B and E D cannot meet, be- 

 caufe they are parallel, and confequently the 

 given Solution cannot take place j for which 

 reafbn I here fubjoin another. 



Another Solution* 

 Upon B C {Tig. 7 .J defcribe a fegment BAC 

 of a circle, fo that the angle of the fegment 

 may be equal to the obferved L£*y, (which 

 as above quoted is fhewn 33. 3. Euclid.) and 

 upon CD defcribe a fegment CE D of a circle 

 capable of an angle equal to the obferved CED; 

 from c draw the diameters of thefe circles cG 

 CH\ then upon c G defcribe a fegment of a 

 circle G FC ? capable of an angle equal to the 

 obferved L aEC\ likewife upon c //defcribe 

 a circle's fegment c F H, capable of an angle 

 equal to the obferved c aE\ fuppofe F the 

 common Se&ion of the two laft circles HFc* 

 GFC, join FHj cutting the circle HE C in E y 

 join alfo FG, cutting the circle G A C in A : 1 

 fay that ^, £, are the points required. 

 Demonflration. 

 " For the Lb a c is— ,3 a y by conftruftion of 

 the fegment, alfo the angles c E c aG, 

 are right, becaufe each exifts in a femicircle : 

 therefore a circle being defcribed upon c F as 

 a diameter, will pafs through £, A • There- 

 fore the angle C A E=L c F E—C p H^(by 

 conftrudion^ to the obferved angle y * V In 

 like manner the L c E A=c F a^cFG— ob- 

 ferv'd angle y s 



In the Rations £, fall in a right line with 

 the point C; the lines G A^ HE being paral- 

 lel. 



