of the Twelfth Axiom of Euclid. 



267 



Fig. 7. 



would be equal and obtuse. Impossible. 

 We have the same result if we suppose 

 BAC, ACD (fig. 6) both obtuse. 



Suppose (fig. 7) one angle B A C to be ob- 

 tuse, and the other ACD acute. From the 

 middle point E of the side A C lying between 

 the unequal angles, draw to B A prolonged 

 a perpendicular E F. E F prolonged from 

 E will meet C D at some point G. The sum 

 of F A E, E A B would be less than the sum 

 of E C G, E A B j therefore E C G would be 

 greater than F AE, and EG longer than EF. 

 Of E G take a part E H equal to E F ; draw 

 CH. C HE wouldbe a right angle, CGH 

 acute, and HGD obtuse. You would 

 then have F, B, D right angles, and F G D 

 obtuse. Impossible. Therefore &c. 



Cor. 2. In a quadrangle having three 

 right angles, if a line is drawn perpendicular 

 to one of the sides which is limit of two right angles from the 

 side opposite to it, the line so drawn cannot be shorter than the 

 other line which is limit of two right angles, nor longer than 

 the one opposite to it. 



In the quadrangle ABCD (fig. 8), let 

 B, D, C be right angles ; from any point E 

 between A and B draw to CD a perpen- 

 dicular E F. If E F were shorter than B D, 

 FEB would be obtuse (Obs.) \ demonstrated 

 impossible. If E F were longer than A C, 

 as F E A must be an obtuse or right angle, 

 E A C would be obtuse. Impossible. There- 

 fore &c. 



Lemma. 



If two right-angled triangles have 

 the hypothenuses equal, and a given 

 acute angle of one be greater than a 

 given acute angle of the other, the 

 second acute angle of the first-named 

 triangle will be smaller than the 

 second acute angle of the other, and 

 all the sides have a relative propor- 

 tion to the angles opposite to them. 



Put one triangle ABC (fig. 9) on 

 the other, A'B'C, so that the right 

 angles may coincide, and the side 

 A C limit of the greater given 

 be on the side A' C limit 



Fig. 8. 



Fig 9. 



angle 



