Properties of Plane Triangles. 33 



sides are the diagonals of tlie rhombi JFZG, dGkl^, d^m F; 

 and consequently the other diagonals dl, dm, dk, also bisecting 

 them, pass through the points K, Q, R. 



Secondly, The sides of the triangle A-/ ?« will pass through 

 I, Y, N and be bisected there. 



For, KI being perpendicular to AC, it is parallel to GA- and 

 Fm, and it is also equal to them. Hence klni is one straight 

 line, and it is bisected in I. 



Cor. 1. — The side of the hexagon E?kF/G/?: is equal to the 

 diameter of the circle circumsciibing the triangle ABC. 



Cor. 2. — The lines le,eP coincide in direction, both pass- 

 ing through e and being perpendicular to AC. Hence the 

 radius of the uiscribed circle P^ passes through /; as do like- 

 wise the radii Zc, O^ through k and 7«. 



Cor. 3. — Let the triangle fclm meet the circle ABC again 

 in p, o, n; then kd, Id, md will pass through these points. 



For Id is perpendicular to GF, and GF is parallel to km ; 

 hence Id is perpendicular to km. In like manner m d is shown 

 to be perpendicular to kl, and kd to Im. Hence Id, md, kd 

 are the three perpendiculars from the angles k, I, m, upon the 

 opposite sides of the triangle klm. But the circle ABC bi- 

 sects the sides km,ml, Ik of this triangle (Pr. X. part 2); and 

 therefore (by Pr. III.) it passes through the feet of the perpen- 

 diculars ; or, Id, md, kd pass through a, n,p>. 



Prop. XL (fig. 2.) 



Join op,pn, no. These lines shall form a triangle whose 

 sides are respectively pai'allel and equal to those of the primi- 

 tive triangle ABC. 



Because KoIB is a rectangle (Ko and IB being perpendi- 

 cular to the two parallels GF, km), the two sides KB, lo are 

 equal. Also, because KF is equal to k\, the two remainders 

 BF and ko are also equal to one another. 



Similarly we find kn equal to FC. And because the angle 

 BFC (or GFE) is equal to the angle okp (or mki), the side 

 o 11 is equal to the side BC. 



Again, because n o, BC make equal angles with the parallel 

 lines kin, GF, they are parallel to one another. In like man- 

 ner it will appear that o^ is parallel and equal to AB, and 

 np to AB. 



Prop. XII. (fig. 2.) 



The point d is the centre of the circle inscribed in the tri- 

 angle 11 o p. 



For n o being parallel to BC, op to AC, and o / to BE, 

 the angle n o I is equal to EBC and p o I to ABE. But the 

 angle EBC is equal to ABE, and therefore no I is equal to 



New Scries. Vol. 2. No. 7. Juli/ 1827. F pol; 



