OF ARTS AND SCIENCES. 



133 



(; DF, FDG, n J) E is not greater tlian R-\-^^RJ^ /; _j- ^^ R = 

 f; J R ; consequently the angle D of the triangle E D G \s not less 

 than 4 /J — ^^ R = ^^ R = 2 R -\- rl'^, which is impossible; hence 

 the sum of the angles of any triangle, A B C.,\s greater than ii-f-Zg R, 

 as required. 



" Prop. 4. The sum of the angles of any triangle is not less than 

 two right angles. 



" If the sum of the angles of any one triangle is not the same as 

 that of any other, then there are evidently some triangles having the 

 sum of their angles less than that of any others ; lot, therefore, ABC 

 (see fig., Prop. 1) denote a triangle* such that the sum of its angles is 

 not greater than that of any other triangle. If jR represents a right 

 angle, then, since the sum of the an- 

 gles of any triangle is greater than 

 12, the sum of the angles of the tri- 

 angle ABC may be expressed by 

 R + V, V being a positive angle. 

 If we denote the angle BAC o[ 

 the given triangle by A, then, as in 

 Prop. 2, we may find some positive 

 integer, m, such that the inequality 

 V ^ 2"i, shall have place ; and by 

 Cor. 1 and 2 of Prop. 1 we may 

 derive from the triangle ABC an- 

 other triangle represented hy D E F 

 such that the sum of its angles shall 

 equal that of ABC, and conse- 

 quently equal R-\-V, and further 



such that the sum of two of its angles, E D F, E F D, shall not be 

 greater than ^- ; .-. the sum of these (two) angles is less than F, con- 

 sequently the third angle E of the triangle is greater than R. Of the 

 two sides D JS, E F, let D E he that which is not the less, and 

 through £ draw JS jfiT, at right angles to DE; then, since the angle 

 D E F IS greater than R, the perpendicular will of course meet the 

 side D F at some point, as K, between D and F. Since the sum of 

 the sides D E and E F is greater than D F (Sim., p. 20, B. I.), and 

 that D K is greater than D E (Sim., p. 19, 13. I.), we of course have 

 D K greater than K F. Hence extend D F to G, so that GK=DK, 

 and extend E K to H, so that KH^— E K, and connect G and // 



