Commentary on Proposition II.11¶
Back to Proposition II.11.
As the solution of this problem is necessary to that of inscribing a regular pentagon in a circle (Eucl. IV. 10, 11), we must necessarily conclude that it was solved by the Pythagoreans, or, in other words, that they discovered the geometrical solution of the quadratic equation [ a(a - x) = x^2, ] or [ x^2 + ax = a^2. ]
The solution in II. 11, too, exactly corresponds to the solution of the more general equation [ x^2 + ax = b^2, ] which, as shown above (pp. 387—8), Simson based upon II. 6. Only Simson’s solution, if applied here, gives us the point F on CA produced and does not directly find the point H. It takes E the middle point of CA, draws AB at right angles to CA and of length equal to CA, and then describes a circle with EB as radius cutting EA produced in F. The only difference between the solution in this case and in the more general case is that AB is here equal to CA instead of being equal to another given straight line $b$.
As in the more general case, there is, from Euclid’s point of view, only one solution.
The construction shows that CF is also divided at A in the manner described in the enunciation, since the rectangle CF, FA is equal to the square on CA.
The problem in II. 11 reappears in VI. 30 in the form of cutting a given straight line in extreme and mean ratio.