314 Mr Meikle on the Theory of Parallel Lines. 



The like evidently follows, when, on the extension of any side of A B C, 

 as, for instance, on that of A C produced to F;, an addition B C F equal to 

 one or more of the foresaid parts of A B C (which have their areas pro- 

 portional to their angular defects) is made to the area ABC: so that, 

 area A B F will likewise be proportional to its angular defect. Hence, 

 in every triangle, the area and the angular defect follow the same pro- 

 portion* 



CaM 2. In the same way, it may be shewn that, if one triangle had 

 the sum of its angles greater than 180°, so would every other, and that 

 the excess (as in spherics) would be proportional to the area. 



Scholium. Since it could still more readily be shewn, either as above 

 or in several other ways, that if one triangle had the sum equal to 180°, 

 80 would every other ; it is obvious that, in respect of 180°, every triangle 

 has the sum of its angles of the same kind, whatever that may be. But 

 there are several known methods of proving that the sum can never 

 exceed 180°. The grand difficulty has always been equivalent to prov- 

 ing that it can never be less. Thus, since, however small might be the 

 amount of some two angles of a triangle, we may always increase those 

 two to be ever so little short of 180°, and yet, according to Euclid's 12th 

 axiom, the sides which have been thereby altered, if produced, will con- 

 tinue to meet, forming a still greater triangle; it is evident that his 

 axiom is equivalent to assuming that the three angles of even the great- 

 est of triangles cannot be sensibly less than 180°. From this second pro- 

 position, too, would follow the converse of the first one, if ever the sum 

 differed from 180°. 



Prop. III. The three angles of every triangle are equal to two right 

 angles. 



Fig. 5 



Let A B C be any equilateral triangle ; 

 bisect its angles, which obviously will 

 divide it into three equal triangles, each 

 having an angle of 120° at the point D. 

 Produce DA, DC to E and F, making 



^ -^ L^ -^ AE = CF rr:^ AC; join AF. Then 



^^- 1 ^"^^ each of the angles A C F, E A F being ex- 



terior in respect of A D C, exceeds this obtuse angle. Hence A F 

 exceeds A C, whilst EF being greater than A F, is still greater than A C 

 or twice A E. Produce B D to bisect A C in G, and E F in H. Let the 

 lines K L, KM, each of which is equal A E, form an angle equal to half 

 the least angle, which any equilateral triangle can ever have : for as long 

 as no two sides of an equilateral triangle can coincide in one straight 

 line, its angle must have some magnitude, and therefore so must the area 



♦ Although, strictly speaking, such reasoning is only applicable to commensur- 

 able areas, it is more than sufficiently general and exact for the present purpose. 



