142 • PROCEEDINGS OF THE AMERICAN ACADEMY 



1 , o -DC' DE DCi AC ■, . ,. ^„ 



have, by prop. 3, -g^ = — = n, .-. j^; = jj^, and since DC = 

 AC,wegetDG = A B. If we take C'B' = CB, and draw a right 

 line from D to B' , the triangles A B C, D B'C are identical, Sim., 

 B. I., p. 4, .•. D B' =AB, and of course D B' =^ D G, which can- 

 not be unless B' falls on G ; for of the base-angles D B'G, D G B\ 

 one is acute and the other obtuse, and the same holds true whether 

 B' is within or without the triangle D F E \ hence we cannot have 

 D B' =^ D G unless B' coincides with Cr, Sim., B. I., p. 19. Hence, 

 since the triangles ABC, D B'C are identical, and that B' falls on 

 G, the triangles A B C, D G C' are identical, and the angles B A C, 

 F D E are equal, as required, 



" Cor. If we draw A K at right angles to B'C\ meeting it in K 

 (see fig., prop. 3), then since the base-angles of the isosceles triangles 

 represented by A B C, A B'C are equal, we have, by cor. to prop. 3, 

 m=^ -r^=^ tttj and hence the vertical angles of the isosceles tri- 



AB AB'^ ° 



angles are equal ; and the same holds true whether the perpendiculars 

 AI, A K fall within the triangles (as in the figure) or without them ; 

 for when the perpendiculars fall without the triangles, the equality 

 -jg- = -jzr, shows that the supplements of the vertical angles of the 

 triangles are equal, and consequently the vertical angles are equal ; 

 and it is evident, by cor. to prop. 3, that the perpendiculars will both 

 at the same time fall within or without the triangles ; the case when 

 Tra = 0, or JB / = 0, B'K = 0, is too evident to require any expla- 

 nation, for the vertical angles are evidently right. Hence the angles 

 A B E, A B'E', the halves of the vertical angles of the isosceles 

 triangles, are equal ; hence it follows that all those right-angled tri- 

 angles which have an acute angle common, or equal, have their 

 other acute angles also equal. Hence (see fig. 3) from the right 

 triangles AB E, A BE', having their angles B, B' equal, we get 



BE B'Ei J • AE AEi i i BE B'E' ,i , • .,. 



IF= H-' ^"^ ^^"^^ Tb = ab!'^^ '^^^"^^ ae=ae'^ that is, if 

 we have two (or more) right triangles which have an acute angle 

 common or equal, then if we divide the leg of any one of them which 

 is opposite to the (equal) angle by the leg adjacent to the angle, the 

 quotient will equal the corresponding quotient obtained in like manner 

 from any other (one) of the triangles ; and the converse of what is 

 here affirmed is also easily proved to be true in a manner very analo- 

 gous to that given in proving the proposition above. 



" Prop. 5. The sum of the acute angles of a right triangle equals 

 a right angle, and the sum of the squares of the legs equals the square 

 of the hypothenuse. 



