The radius of curvature of a curve $\gamma:I \to \mathbf{R}$ parametric by arc length is $||\ddot \gamma||^{-1}$.
I want to demonstrate this using synthetic geometry.
Let $A$, $B$ and $C$ be three noncolinear points with $AB = BC = 1$ and let $O$ be the center of the circle passing through $A$, $B$ and $C$. And let $D$ be a point such that $ABCD$ is a parallelogram. This is illustrated below:
I want to show that
$$BD \times BO = 1.$$
This is simple enough to do with trigonometry. Bur is there a more elegant way to show this?