472 PROFESSOR KELLAND ON SUPERPOSITION. 



XIV. BG and BY are the same as in the last method, whilst a parallel is drawn 

 from Y instead of from G. 



(1), (2), and (3) will nnite as in the second figure. Also, since the sides of (3) 

 and (4) are a and x—a, the square is complete (by No. XIII.) ; hence the con- 

 clusion. 



XV. (1) and (3) are the same as in method XIII : (4) is constructed by draw- 

 ing GH equal and parallel to AY, and completing the rectangle. (1), (2), and (3 ) 

 will unite as in the second figure ; and the conclusion is effected as in XIV. 



XVI. Cut off BY, BV, each equal to x., draw VE parallel to BY, join CY and 

 produce it to meet VE in D. 



(1), (2), and (3) will unite, as m the second figure ; and the length of (4) is the 

 same as that of (3) ; hence the conclusion. 



The division is, in this case, effected by two cuts. 



Cor. A modification of this method may be produced by omitting CD, and 

 a triangle equal to CDE out of (3), as in the dotted line. 



XVII. Make BT = ^; draw TQ parallel to AB; make CR = DQ, and draw FG 

 parallel to AQ through any point F, within the limits indicated in the figure by 

 cutting the points C and T ; draw RH parallel to AB. 



(1), (2), (3), and (4) will unite as in the second figure ; hence the conclusion. 



XVIII. Bisect PC in Q. With centre B and radius BR = a?, describe a circle ; 

 and from Q, draw QR touching the circle in R ; produce QR to meet BA in S ; 

 make BT = BS, and draw TV parallel to BR meeting DE in V. 



(1), (2), and (3), will unite as in the second figure, and the side of (4) will be in 

 the same straight line with the sides of (1) and (2) ; hence the conclusion. 



XIX. Bisect CD in Q. With centre A and radius AR = -a?, describe a circle ; 

 and from Q draw QR touching the circle in R ; produce QR to meet BE in H ; 

 make ET=r«, and draw TL parallel to AB. 



(1) and (3) unite as in the second figure, so that the longest side of the united 

 figure is 2«, consequently (2) falls upon it ; hence the conclusion. 

 The division may, in this instance, be effected by two cuts. 



XX. Bisect CD in F, draw FG parallel to AB ; make AH = AY = 2«— a; ; join 

 AH, and draw YZ parallel to BE. 



The triangle is HGK equal to AYZ, hence (1), (2), (3), and (4) will unite, as in 

 the second figure, and the figure is a square. 



XXI. Bisect CD in G ; draw GH parallel to AB ; draw BJ, making an angle of 

 30° with BH, and AK perpendicular to BJ. 



(1), (2), and (4) will unite as in the second figure, and AK and BJ are each 

 equal to x ; hence the conclusion. 



XXII. Bisect CD in G, and through G draw GH parallel to AB, meeting AP 

 in J. Draw GL, making an angle of 30° with CD, and make LO = a. Through 



