266 Mr. B. A. Murray on a rigorous Demonstration 



pendicular fall at a point G (fig. 4) between C and E, and being- 

 prolonged from G meet A B at a point H between A and B, from 



Fig. 4. 



H prolong F H until its prolongation H I is equal to F H. From 

 I draw a perpendicular to H A, and suppose it to fall at a point 

 K between A and II. The triangles B F H, K I H are equal, the 

 sides B II, K H are equal ; and as H B is longer than GE, KH 

 is longer than G E ; and considering the quadrangle HBGD, 

 the entire quantity K H is in advance of GE (Obs.). And sup- 

 posing I K prolonged from K to meet CD ata point L between 

 C and G, L G would be longer than K H. Now by prolonging 

 I L, as in the preceding cases, you can construct a triangle equal 

 and similar to GIL, whose perpendicular will cut off D C, or its 

 prolongation, a quantity (equal to G L) in advance of and greater 

 than K H, and, a fortiori, greater than all preceding similar 

 smaller quantities as G E. Now supposing the sides A B, D 

 to be not more than m times as long as G E, by constructing m 

 triangles with the conditions of construction of E F G, the per- 

 pendicular of one of them must, if prolonged, meet the side AC 

 at C or E as in fig. 1, or meet the extremity (A or C) of A B or 

 C D as in fig. 2, or meet the prolongation of one of them as in 

 fig. 3, thus demonstrating, as in those cases, the acuteness and 

 equality of the angles ACD, B A C. Therefore &c. 



Cor. I. In a quadrangle, if two adjacent angles are right an- 

 gles, the sum of the others cannot exceed two right angles. 



It is demonstrated to be impossible if the 

 opposite sides comprising the two adjacent 

 right angles are equal. Suppose those sides 

 to be unequal, and in the quadrangle 

 ABCD (tig. 5) the angle BAG to be 

 a right angle and ACD obtuse. Prolong 

 from C the shortest side D C {Obs.) until it, 

 I) E, is equal to BA; the angles BAE, AED 



Fie. 5- 



