312 



Mr Meikle on the Theory of Parallel Lines. 



this subject. For, so far as I am aware, it had not till now been shewn 

 that, if, in so much as one triangle, the sum of the angles differed from 

 180°, a definite relation behoved to subsist between the areas and angles 

 of all triangles. It is to be hoped that such a relation, which I have 

 here deduced from that supposition, and endeavoured to follow it up 

 to an absurdity, may yet lead to other and preferable modes of de- 

 monstration. 



Pkop. I. Triangles, whose areas are equal, have the sums of their 

 angles equal. 



Fig. 1. Fig. 2. 



B N Q 



Case 1. To prove this of any two triangles, A B C, D E F, which have 

 their bases equal, as well as their areas ; bisect A B, B C in G and H, 

 and to GH or its extension draw the perpendiculars A I, BK, CL. 

 Then the triangles A 1 G, B K G, having two angles and a corresponding 

 side respectively equal, are (Eucl. i. 26) every way equal ; and for the 

 like reason, the triangles B K H, C L H, are equal. Hence, the triangle 

 A B C is equal to the quadrilateral A I L C, and its three angles are 

 equal to the two angles I A C, ACL. 



The three perpendiculars drawn to I L have in effect been proved to 

 be equal ; and in the same way as it has been shewn of ABC, every 

 other triangle which could have A C for a base, and whose sides would be 

 bisected by G H produced, if necessary, must have its perpendiculars 

 which are drawn to this bisecting line, as well as its area, and sum of angles 

 respectively, equal to those of ABC. Bisect, therefore, I L in M, and 

 on it erect the perpendicular M N =i A I ; join AN, N C, cutting I L 

 in O and P. Then, nearly in the same way as above, the triangles AID, 

 N M O, N M P, CLP, being found equal, first in pairs, and then all 

 four ; the triangle A N C is isosceles, and has its area equal to A I L C = 

 ABC, and its three angles equal to the same two angles as before, or 

 to the three angles of A B C. 



In like manner, construct an isosceles triangle D Q F = D E F. Then 

 the areas and bases of the two isosceles triangles A N C, D Q F, being 

 respectively equal, there is no alternative but their angles must agree ; 

 for otherwise, a copy of the one triangle being constructed on the base 

 of the other, their areas behoved to differ. Consequently, the triangles 

 ABC, D E F, have the sums of their angles equal. 



