396 
EUCLID 
order to construct the required gnomon, we have only to draw 
in the angle FGE the parallelogram GOQP equal to (EF) — C 
and similar and similarly situated to D. This is what Euclid 
in fact does; he constructs the parallelogram LKNM equal to 
(EF) — C and similar and similarly situated to D (by means of 
25), and then draws GOQP equal to it. The problem is thus 
solved, TASQ being the required parallelogram. 
To show the correspondence to the solution of a quadratic 
equation, let AB — a, QS = x, and let b:c be the ratio of the 
sides of D; therefore SB =-x. Then, if m is a certain con- 
c 
stant (in fact the sine of an angle of one of the parallelograms), 
(AQ) = m {ax — ~x 2 ), so that the equation solved is 
Euclid gives only one solution (that corresponding to the 
negative sign), but he was of course aware that there are two, 
and how he could exhibit the second in the figure. 
For a real solution we must have C not greater than 
C Ct‘ 
m b * ~4 ’ w ^i c h i' s the ares, of EF. This corresponds to Pro 
position 27. 
We observe that what Euclid in fact does is to find the 
parallelogram GOQP which is of given shape (namely such 
that its area m. GO .OQ = m. GO 2 -) and is equal to (EF) — G; 
c 
c /c a ^ Cl \ 
that is, he finds GO such that GO 2 = t(t* )• In other 
b\b 4 m/ 
b \b 4 
words, he finds the straight line equal to ^ J ~ —; 
and x is thus known, since x = GE — GO — GO. 
b 2 
Euclid’s procedure, therefore, corresponds closely to the alge 
braic solution. 
The solution of 29 is exactly similar, mutatis mutandis. 
A solution is always possible, so that no Siopar/xos is required.