180 PROCEEDINGS OF THE [Sept, 1887. 



each pair of external segments about any angle are equal to each other, and the 

 three bisectrices AK, BK, CK, have a common point K. 



5. Prop. III. The segments BL and CH are equal, and consequently, also 

 BH and CL are equal. 



First, since AI=AM, and AE=AB, we have EI=FM ; secondly, since BI 

 and BE equal respectively BL and BH, we have EI=BL+BH=BL+BL-t- 

 LH ; thirdly, in like manner, since CF and CM equal respectively CH and CL, 

 we have FM=CH+ CL=CH+CH+LH ; fourthly, by combining these results, 

 we get BL+BL+LH = CH-|-CH+LH which shows that BL=CH, and there- 

 fore also that BH=CL. 



6. Pnop. IV. First, the line AI equals half the j^erimeter of the triangle 

 ABC ; secondly, BI is the difference between the half perimeter and the side 

 A15 ; thii'dly, UE is the difference between the half perimeter and side AC ; 

 fourthly, AE is the difference between the half periiueter and side BC. 



First, the perimeter is composed of the six segments already mentioned, for- 

 ming three pairs, the members of each pair being equal to each other. Then if 

 one member be taken from each pair, their sum will equal the half perimeter, 

 and such is the constitution of the line A I, for AE is one of the pair at angle 

 A, BE is one of the pair at angle B, and BI is equal to one of the pair at angle C. 



Secondly, it is obvious that BI is the difference between the half perimeter 

 AI and the side AB. 



Thirdly, since BI=CF and AE=AF, so that AE+BIrrCF+AF=side AC, 

 therefore BE is the difference between the half perimeter AI and side AC. 



Fourthly, since BE=BH and BI =rCH, then EI=BC, and AE is the differ- 

 ence between the half perimeter AI and the side BC/. 



7. Prop. V. The right angled triangles BDE and BKI are similar, and the 

 rectangle BExBI=the rectangle DExKI. 



The angle CBI is the supplement of CBA, and therefore the angle KB I, 

 which is half the fii'st, must be the complement of DBE which is half the sec- 

 ond. But EDB is also the complement of DBE, and is therefore equal to KBI. 

 Then the triangles having two angles in one respectively equal to two angles in 

 the other are similar, and, DE : EB::BI : KI and the rectangle BExBt= 

 the rectangle DExKI. 



8. Prop. VI. The area of ABC equals the rectangle AIxED. 



For the area of ABC is equal to the sum of the areas of the triangles ADD, 

 BCD and CAD, which have altitudes equal to ED. Hence the area ABC 

 equals half the sum of the bases of these triangles upon their common altitude 

 and therefore equals half the perimeter upon the common altitude, or area 

 ABC equal the rectangle AIxED. 



9. Prop. VII, The area of triangle ABC is a mean proportional between 

 the areas of two rectangles, the sides of the one being AE and AI, those of the 

 other BE and BI. 



The triangles AED and AIK are similar, and we get AE : ED :: AI : IK. 



Takiag first and second terms of the proportion as bases, form rectangles by 

 using third term as altitude ; taking third and foui-th terms as bases form rect- 



