If we take a geometric view of Theon's *side* and *diagonal* numbers, we can literally picture them as the sides \(x\) and diagonal \(y\) of an isosceles (but non-right) triangle. The resulting triangles will alternate between acute and obtuse as indicated by \(\pm\) in the equation \(y^2=2x^2 \pm1.\) See Figures 2 and 3.

The area of the square on the base, \(y^2,\) and the sum of the squares on the legs, \(2x^2,\) always differ by exactly 1. Since the areas are increasing, it must be that \(y^2\) approaches \(2x^2.\) Visually, the alternating acute/obtuse angle \(\theta_n\) is approaching a right angle. We will assume for the remainder of this section that \(\displaystyle \lim_{n\rightarrow \infty}\theta_n\) exists (so we can use the limit rules) and that the limit is equal to \(\displaystyle \frac{\pi}{2}.\) The sequence of triangles is approaching a right isosceles triangle, and hence the ratio of the diagonal number to the side number is approaching \(\sqrt{2}.\)

**Figure 2.** The case \(x=1, y=1\) with areas of squares

**Figure 3.** The case \(x=2, y=3\) with areas of squares

So, as the iterative process continues, \(\cos{\theta_n}\) approaches zero. The following animation (of a GeoGebra applet) illustrates convergence as \(n\) increases.

The applet itself is available here.

Let's consider what happens if we look at this visualization in terms of trigonometric functions. Trigonometric functions were not in use in Theon's time, but we can use them to provide an alternate proof of Theon's method. In Figure 4, we note that since the triangle \(ABC\) is isosceles, the angle bisector is a perpendicular bisector of the base \(BC.\) From right triangle trigonometry, \(DC=x\sin(\theta/2).\) Thus the length of \(y=BC\) is \(2x\sin(\theta/2).\) We can check that as \(\theta\) approaches a right angle, the ratio \(\displaystyle\frac{y}{x}\) approaches our expected \(\sqrt{2}.\) That is, \[\frac{y}{x} = \frac{2x\sin(\theta/2)}{x} \rightarrow 2\sin\left(\frac{\pi}{4}\right) = \sqrt{2},\,\,\,{\rm{as}}\,\,\, \theta \rightarrow\frac{\pi}{2}.\]

**Figure 4.** The side opposite angle \(\theta\) has length \(y=2x\sin(\theta/2).\)

Further, if we consider the Law of Cosines applied to the isosceles triangle with legs of length \(x,\) base \(y,\) and vertex angle \(\theta,\) we have \[y^2=x^2+x^2-2x\cdot x\cos\theta,\] which simplifies to \[y^2=2x^2(1-\cos\theta).\] From this we see the ratio \[\frac{y^2}{x^2}=2(1-\cos\theta).\] And, as assumed, as \(\theta\) approaches a right angle, \(\displaystyle\frac{y^2}{x^2}\) approaches \(2.\) That is, \(\displaystyle\frac{y}{x}\) approaches \(\sqrt{2}.\)

As a little bonus, substituting the result from Figure 4 into \(y^2=2x^2(1-\cos\theta),\) we obtain \[4x^2\sin^2(\theta/2)=2x^2(1-\cos\theta).\] Assuming \(x\neq0,\) we verify the half-angle formula, \[2\sin^2(\theta/2)=1-\cos\theta.\]