no GEOMETRICAL PORISMS. 



PROP. II. POR ISM, Fig. 7. PI. I. 



Let AF, AG be two Uraight lines given by pofition, a point 

 H may be found, fuch, that any circle whatfoever defcri- 

 bed tlirough it, and A the interfeiflion of the given lines, 

 to meet them in D and E, fhall cut off from them feg- 

 ments AD, AE, whofe fum fhali be a given line M. 



Suppose the porifm to be true, and that the point is found, 

 and circle defcribed as above, let given points B, C be fo taken, 

 that BA and AC may be together equal to DA and AE, that is, 

 by hypothefis to the given line M, then BD vsrill be equal to CE. 

 If a circle be defcribed through the given points A, B, C, by hy- 

 pothefis it will meet the circle palling through A, D, E, in H the 

 point which may be found. Join BH, CH, DH, EH. The 

 angle BHC is equal to DHE, each being the fupplement of BAG, 

 therefore BHD is equal to CHE; now, HDB is equal to HEC, 

 and BD is equal to CE, therefore the triangle HBD is equal to 

 HCE, and BH is equal to CH, alfo DH to EH ; hence the angle 

 BAH is equal to CAH, and H is in a llraight line bifedling the 

 angle FAG, but it is alfo in the given circle BAG ; therefore the 

 point H is given, as was required. 



Hence this conflrudlion : Take B and C two given points, i^a 

 that BA and AC may be togetlier equal to M, and through 

 A, B, G defcribe a circle. Draw AK bifecfting the given angle 

 FAG, and meeting the circle ABC in H the point required, that 

 is, if any circle be defcribed through H and A, to meet the gi- 

 ven lines in D and E, the fum of DA and AE fliall be equal to 

 the fum of BA aiid AC, that is, by conftru(flion to the given 

 line M. The fynthetical demonftration follows readily from 

 the preceding analyfis. 



PROP. III. 



