of the Twelfth Axiom of Euclid. 



269 



G H, A B of the quadrangle ABGH, perpendiculars A I, 6K. 

 With A I for ray describe a circle I Q, with AC describe C N, 

 and with A L (L the intersection of A E and C F) describe L M. 

 From O, intersection of G A and circle C N, draw to A B a per- 

 pendicular P ; prolong it from till it meets the circle I Q at R; 

 draw A R. From M and S, intersections of A R and circles L M 

 and C N, draw to A B perpendiculars MT,SU; from A toV, in-, 

 tersection of circle I Q and G K, draw AV. From W, intersection 

 of A G and circle I Q, and from X, intersection A V and circle 

 C N, draw perpendiculars WZ,XY. From M, L, S, 0, and X 

 draw to S U, R P, W Z, and V K perpendiculars Ma, Lb, Sc, 

 O d, and X e. The right-angled triangles A S II, A O P have 

 the hypothenuses A S, A equal. The angle A P >■ A S U 

 (Lem.), FP>TU (Cor. 2, first principal proposition) ; F U is 

 common to T U and F P ; .\ U P > T F. 



The right-angled triangles ARP, AWZ, AVK have the hy- 

 pothenuses equal ; the angle AVK>AWZ>ARP (Lem.) ■ 

 therefore in the right-angled triangles S R c, W d, XVe (the 

 hypothenuses SR, OW, XV being equal), the side Xe>0*/^Sc; 

 therefore the segment YK>PZ>UP (Cor. 2, first principal 

 proposition) ; and as we have shown U P > T F, .". YK > T F ; 

 that is, the perpendicular G K in the quadrangle ABGH cuts 

 off the side A B a quantity Y K similar to, but greater than and 

 in advance of T F cut off by the homologous perpendicular C F 

 in the quadrangle ABCD. Now, supposing the sides A B, C D 

 to be not more than m times as long as T F, by constructing m 

 triangles with the conditions of construction of ACF, AGK, &c, 

 a perpendicular of one of them, homologous to C F or G K, will 



fall on the prolongation of the side, 

 AB (or CD, as may be) ; and if 

 the angles A B D, B JD C were 

 right angles, there would be two 

 perpendiculars drawn from one 

 point to a straight line, which is 

 impossible. The angles ABD, 

 BBC therefore cannot be right 

 angles. Suppose them to be acute, 

 and A C, B J) to be equal ; from 

 the middle point of AC to the 

 middle point of B D (fig. 1 1 ) draw 

 a line EF ; it will be perpendicular 

 to both («), and you would have 

 a construction (two right angles 

 and two equal and acute) just de- 

 monstrated impossible. Suppose 

 A C longer than B D. Take E A', 



Fig. 11. 



