# Oops, Square roots!

It was pointed out yesterday that I forgot to prove, or even mention the fact that $\mathbb K$ is closed under taking square roots. That is, if $k\in\mathbb K$, then so is $\sqrt k$. So here goes.

Suppose we can construct $k$. Build a line segment $AB$ of length $k$. Construct an additional segment of length 1 after it, so we have a segment $AC$ of length $k+1$. Now draw a circle with this as the diameter. You should have something like this:

Great! Now we can drop a perpendicular at $B$, and see where it intersects the circle. Call this point $D$. Remember from my first post on constructibility that we can construct perpendiculars through a point. We now have:

Next we draw in lines $AD$ and $CD$, and notice that the triangle $\Delta ADC$ is a right triangle.

We can then call on some altitude on hypotenuse theorems. Namely, $\frac{AB}{BD}=\frac{BD}{BC}$, so $BD=\sqrt{AB\cdot BC}$ But in our diagram, $AB=k$, and $BC=1$, so $BD=\sqrt k$, as desired.

In the last few lines of text, I think you mean $BD = \sqrt{AB \cdot BC}$ (instead of $BC = \sqrt{AB \cdot BC}$) and $BD = \sqrt{k}$ (instead of $BC = \sqrt{k}$, since $BC=1$).