Properties of Plane Triangles. 29 



< AP.CP "^ AP. BO "*■ CP. BO 1 '^ ~" ^'-'^'' + ^"^" '' "^ '"' • ^'■ 



= S" (as before) ; 



AP. CP ' AP. BO ' CP. BO 

 1 1 1 



' AP. CP ' AP.BO ' CP.BO r> 



The last three I believe are original ; and it is obvious that 

 a similar mode of proceeding is calculated to develop a consi- 

 derable number of neat properties, as well as to demonstrate 

 very concisely several of those already known. 



But I must proceed to the main object of my paper. 



Prop. I. Plate I. (fig. 1.) 



Let ABC be any plane triangle, and let perpendiculars 

 A^, BF, Ca be demitted from each angle upon its opposite 

 side, and prolonged to meet the circumscribing circle in /c,h, i: 

 then if the triangle k h i be formed, its angles will be bisected 

 by the said perpendiculars. 



For, since the lines C«, A6 are perpendicular to the lines 

 AB, BC, and the angle ABC common to the two triangles 

 AZ»B, CaB, the angle BA6 is equal to the angle BC«, 

 Hence they stand on equal arcs f B, BX". But the angles Vthi, 

 B/t k stand on the same two arcs, and therefore they are equal ; 

 or the angle ikh is bisected by B//. 



In the same manner, the angles hik^ ikh are proved to be 

 bisected by Ca, and Kh respectively. 



Cor. 1. — The angles of the triangle ahY are also bisected 

 by the same perpendiculars. For each side of this triangle 

 is manifestly parallel to a corresponding side oi ikh. 



Cor. 2. — Each of the triangles BaZ>, 6CF, and FA«, is si- 

 milar to the original triangle. For the angles a FB, a CB 

 being equal, their complements a FA and «BC are equal. In 

 like manner it may be shown that Fa A is equal to BCA ; and 

 therefore the triangle «FA is similar to ABC. And so of 

 the others. 



Or this corollary may be thus deduced: Because A«C, 

 Ai^C are right angles, a circle will pass through Aa^C ; and, 

 therefore, the angles V>ab and Bia are equal respectively to 

 BCA and BAG. And so of the others. 



Prop. II. (fig. 1.) 

 A circle described through the icet of the perjjendiculars 

 a, iJ*, F will also bisect the sicles of tlie triangle. 



For, let the circle cut OB in /, AO in r, and OC in r ; AC 

 in 1), AB in L, and liC in .1. Also, joiny'I), tL, and «'J. 



Then, 



