OF ARTS AND SCIENCES. 



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any given angle. Hence, since the angles C and B are infinitesi- 

 mal angles, the sides A C and A B must coincide very nearly with 

 the side C B, and .•. since A C and A B lie in opposite directions 

 they cannot possibly come near to coincidence with each other ; but 

 since the angle A is less than any given angle (or infinitesimal), the 

 sides A C, A B must very nearly coincide with each other and have 

 nearly the same direction, which is absurd. Hence the sum of the 

 angles of a triangle is not less than any given angle (or infinitesimal), 

 but it is a finite quantity, being equal to some finite angle, or the sum 

 of finite angles. 



'''■Remark. By aid of this lemma we are prepared to give a very 

 simple demonstration of Prop. 3. 



" Prop. 3'. The sum of the angles of any triangle is greater than a 

 right angle. Let the triangle A B C, o{ Prop. 1, denote any triangle, 

 and denote the angle BAG hy A, and use V to 

 represent any small finite angle ; then we may 

 find some positive integer, m, such that the ine- 

 quality m^'^A shall exist, consequently the in- 

 equality 5^ ^i has place also. Hence, by Cor. 

 1 and 2 of Prop. 1 , we can find a derived trian- 

 gle, which we shall represent by the triangle 

 H IK, such that the sum of its angles equals that 

 of A B C, and the sum of two of its angles, IHK, 

 IKH, is not greater than -1^^, .*. ^ is greater 

 than the sum of these two angles. At the point I 

 make the angle HIL, equal to the angle HIK; 

 also make the right line I L equal to the side IK, and draw a right 

 line from H lo L; then (Sim., p. 4, B. I.) the triangles HIK, HIL 

 are identical, making the side L H equal to the side H K, the angle 

 IH L equal to the angle IHK, and the angle IL H equal to the angle 

 IKH] hence V is greater than the sum of the four angles IHK, 

 I H L, I K H, I L H. If we connect the points K and L by a right 

 line, it will intersect I H, or JiJ produced, in some point, M, and the 

 angles at M will be right angles ; for the triangles KHM, LHM are 

 identical, since H K = H L, and the side HM common, and that 

 the angle KHM equals the angle L HM, .-. the angle KMH equals 

 the angle LMH (Sim., p. 4, B. I.) ; consequently KM is a perpen- 

 dicular from the angle ^of the triangle H I K to the opposite side 

 H I, or to H I produced (Sim., def 10, B. I.). We now observe that 



