Mr Meikle on the Theory of Parallel Lines. 313 



Case 2. When ABC, D E F (figs. 1 & 3) are any two equal triangles 

 which have no equal bases; bisect DE, 

 F E in G, H ; and let A B C be the triangle 

 which has the greatest side; then an arc 

 having D as a centre, and a radius equal to 

 half that side of A B C, will meet G H or its 

 extension in some point I. Produce DI 

 to double its length in K, join F K meeting 

 GH in 0, and to GH draw the perpen- 

 diculars DL, FM, K N. Then, as remarked 

 above, D L == F M, and therefore, the triangles D I L, KIN, having 

 two angles and a corresponding side respectively equal, we have also 

 KN = DL = FM; and since, therefore, the triangles K N 0, F M O, 

 have likewise two angles and a corresponding side equal, L M bisects 

 F K in O. Hence, as noticed under the first case, the triangle D K F 

 has its area and the sum of its angles equal to those of D E F, since 

 their sides are bisected by tlie same line. Again, by the first case, the 

 triangles D K F, ABC, have the sums of their angles equal ; for D K 

 is equal to one side of A B C, and their areas are equal. Consequently, 

 any two triangles ABC, D E F, whose areas are equal, have the sums 

 of their angles equal. 



Prop. II. If, in one triangle, the sum of the angles differed from ISO"* 

 so it would in every triangle ; the difference pig, 4. 



would always have the same sign, and be pro- 

 portional to the area. 



Case 1. Let A B C be a triangle whose angles, 

 if possible, fall short of 180°, and let its area 

 be bisected by B D. Then each of these halves 

 will (Prop. 1.) have the same amount of angles ; 

 but whether they were halves or unequal a e d" 

 parts, their six angles would evidently be equal to the three angles of 

 ABC, together with 180° at the'point D : so that, the six angles of the 

 two parts will always be less than 360° by the same quantity that the 

 three angles of A B C are less than 180^ Consequently, the defect of 

 half the sum of the six from 180° will just be half the defect of the three 

 angles of A B C from 180°. In the same way, if B E bisect the area 

 A B D, the angular defect of each part from 180' will be half of the de- 

 fect for A B D, or one- fourth of that for A B C ; and so on, for the con- 

 tinual bisection of the whole till each part be less than any given area. 



But however unequal the two parts may be into which any one of such 

 bisecting lines divides the whole or a part of the area ABC, the same rela- 

 tion will still subsist. For when the several parts of the areas and of the 

 angular defects are proportional, so must any corresponding sums of such 

 parts. Thus area E B C will be proportional to its angular defect. 



VOL. XXXVI. NO. LXXII. APRIL 1844. X 



