A Kick in the Discovery

As it happens, we want to show that $\angle ch$ is a right angle.

To do so, consider the following figure:

You might notice that we have two isosceles triangles (each has a radius as two of its legs). From this, we deduce: $2\beta+2\theta = 180\degree,$ which implies $\beta+\theta = 90\degree.$

Remark: We can also deduce the following as well: $\alpha+ \phi = 180\degree,\;\;2\theta + \phi = 180\degree.$ From these equations, we can see that $\alpha =2\theta,$ which you may recall comes from the inscribed angle theorem.

In summary, we have:

Perfect, we have the following:

The key step is to notice that $\triangle ahc$ and $\triangle hbd$ are similar triangles (which follows from the fact that they both have the same internal angles, can you see why?). Thus, $a/h = h/b$ from which we deduce $h = \sqrt{ab}.$

The final steps are to note that $h \leq radius$, for any triangle that we draw, and that $radius = \frac{a+b}{2}.$