PROFESSOR KELLAND'S PROBLEM ON SUPERPOSITION. 309 



It is manifest that many of the preceding solutions, for example, Figs. 9, 10 ; 33, 

 34 ; 41, 42, would apply to gnomons where the breadth of one leg is equal to the length 

 of the other. 



The solutions in Figs. 7, 8 ; 13, 14 ; 15, 16 will apply when the legs of the gnomon 

 are equal in breadth. 



On Cutting Rectilineal Figures by Straight Lines into Pieces that shall be capable of 



forming other Rectilineal Figures. 



To cut a parallelogram, ABCD, into another with the same angles, but with a side 

 equal to EF (fig. l). First, when EF is less than one of the sides. Lay off DK, BG, 

 GH, HI, &c, each equal to EF, join AK, and draw parallels. It is manifest the pieces 

 1, 2, 3, 4, 5 will form the required figure. Second, when EF is longer than either of the 

 sides. Find by geometrical construction the other and short side, and proceed as before. 

 Hence to cut a rectangle into a square, (fig. 2), lay off DK, BE equal to side of square, &c, 

 and BEFG is evidently equal to the square. If length of rectangle exceeds four times 

 the breadth, four pieces are required; if exceeding nine times, five pieces are required, 

 and so on. 



II. 



To cut two parallelograms having the same angles into one. Cut one so as to have a 

 side = one of the sides of the other, and place them together. 



Or place the parallelograms ABCD and BEFG as in fig. 3, lay off EK = AB, join DK, 

 KF, draw DH parallel to KF meeting BC produced in H, join HF, make GL = AB and 

 LMllBH. The sides of the triangles DCH, KEF are parallel, and side DC = CK, .*. 

 DH = KF, and . \ KH is a parallelogram, and the five pieces of which it is composed 

 are those of the given parallelograms. N.B. — If DK cut BC between G and C, another 

 piece will be required. 



To cut two squares into one, proceed as in fig. 4 ; and in the same way two similar 

 rectangles may be cut into a similar rectangle. 



III. 



To cut a parallelogram, ABCD, (fig. 5), into another, ABEF, on the same base and of 

 the same height but with a given L BAF, or a given side AF not less than the height of 

 the parallelogram. Lay off L BAF, or insert the line AF, and proceed as in fig. 5. The 

 proof is manifest. N.B. — The above illustrates Euclid, i. 35, by superposition. 



