Dinostratus proved that the trisectrix of Hippias could be used to solve this problem after discovering that the side of the square is the mean proportional between the arc of the quarter circle AC and the segment DQ.

There are various stages to the reductio ad absurdum proof which are illustrated in the following windows:

Let the circle with centre D and radius DR intersect the trisectrix at S and the side of the square at T. Draw the perpendicular SU to side DC from point S. As the arcs are proportional to the radii then AC/AB=TR/DR    (2)

From (1) and (2) it must follow that TR=AB          (3)

S is the point on the trisectrix which satisfies 

TR/SR=AB/SU                                      (4)        

From (3) and (4) it follows that SR=SU

However, this is absurd as the perpendicular is the shortest distance between a point and a line. Therefore, DR cannot be longer than DQ.

2.- We repeat this way of reasoning with the hypothesis R<Q and once again the result is false.

3.- Since AC/AB=AB/DQ, then AB·AB=DQ·AC and AC is the fourth proportional of the three other straight line segments.

In this window you can see the construction of a rectilinear segment which is equal in length to AC. This arc is equal to a quarter of the circumference.

Note that the line passing through A and Q is the diagonal of the rectangle OCPS and hence divides the rectangle into two equal parts. If we compare the two parts we can see that the area of the square ABCD is equal to the area of the rectangle ARST since the triangles are equal to each other.

5.- Since the rectilinear segments and curvilinear segment are both the fourth proportional of the same proportion then they must be equal.

Eudoxus had already demonstrated that the area of a circle was equal to half of its circumference multiplied by its radius through his reductio ad absurdum technique. The area of the rectangle of sides a and 2b, where 2b is the rectilinear segment which coincides with the semicircle, is equal to that of the circle. From this rectangle is it easy to obtain a square of the same area using the proportional mean of a and 2b.

This process allows segments to be constructed which measure the square root of n. It uses segments which measure the root of n-1 and 1 and is based on Pythagoras' theorem.

7.- Note that the ends of the lines are joined together making the diagram look like a spiral.

8.- You can reduce the process starting with the square closest to the number whose square root you want to construct. For example, for 19 start with 16, i.e. with the segment that measures 4 and follow on with 17 and 18 until you get to 19.