2 
Transactions of the Society. 
demonstrated as follows. Let AC = a and BC = 6; from the point 
D, where the thread (dotted line) cuts the diagonal, draw D E and 
D F perpendicular to A C and B C respectively. Call A E, p; 
EC,?/; and B F, s. 
Fig. l. 
/I 
Because the ang’ie D C E is half a right angle, and D E C a right 
angle, therefore CDE is also half a right angle, and consequently 
equal to DCF. Therefore DE=EC=y and D E C F is a square. 
Because A B meets the two parallels D F, A C the angle B D F is equal 
to the angle D A E, consequently their tangents are equal, thus 
£=», and 
y p 
?/ 2 
s = — a = T^v\ 
Therefore 
6 = V + s = y + 
y (p + y) 
V 
- 4- - = 1 I 2 = 1. 
a h p + y y{p + y) y 
In order to find then the sum of — f- - place a T-square against 
A C to find the point E, where the perpendicular D E from the point 
D meets A C ; then E C is the required answer. The scale on 
the diagonal line therefore is not required to obtain a solution, but it 
is obvious that the scale EC may be transferred to C D, then the 
answer can be read off directly at the point D without the use of the 
T-square. For as x=y x / 2 it is only necessary to make the divisions on 
CD 2 times larger than those on A C or B C. For example, if the 
