I (re-)discovered the following result recently.
Inscribed angle theorem for hyperbola
The Minkowski inner product on $\mathbb{R}^{1,1}$, $u\cdot v=u_1v_1-u_2v_2$, satisfies a version of the Cauchy-Schwarz inequality: $(u \cdot v)^2 \geq u^2 v^2$. If $u$ and $v$ are spacelike vectors (i.e. $u^2 >0$ and $v^2>0$) then it is possible to define an 'angle' between them via $\cosh(\theta)=\frac{u \cdot v}{|u||v|}$. In the language of special relativity, $\theta$ is the rapidity between two observers moving along tangent vectors $u$ and $v$.
The analogue of the unit circle in $\mathbb{R}^{1,1}$ is the hyperbola $y^2-x^2=1$. Take 3 points on the upper branch of the hyperbola $P_0=(\sinh(\theta_0),\cosh(\theta_0))$ and $P_1$ and $P_2$ defined similarly.
The gradient of $P_1P_0$ is
$\frac{\cosh(\theta_0)-\cosh(\theta_1)}{\sinh(\theta_0)-\sinh(\theta_1)}=\frac{2\sinh\left(\frac{\theta_0+\theta_1}{2}\right)\sinh\left(\frac{\theta_0-\theta_1}{2}\right)}{2\cosh\left(\frac{\theta_0+\theta_1}{2}\right)\sinh\left(\frac{\theta_0-\theta_1}{2}\right)}=\tanh\left(\frac{1}{2}(\theta_0+\theta_1)\right)$.
So the angle between $P_1P_0$ and $P_2P_0$ is given by
$\cosh(\theta)=\frac{\left( \matrix{1\\\tanh\left(\frac{\theta_0+\theta_1}{2}\right)} \right)\cdot\left( \matrix{1\\\tanh\left(\frac{\theta_0+\theta_2}{2}\right)} \right)}{\sqrt{1-\tanh^2\left(\frac{\theta_0+\theta_1}{2}\right)} \sqrt{1-\tanh^2\left(\frac{\theta_0+\theta_2}{2}\right)}}=\cosh\left(\frac{\theta_0+\theta_1}{2}\right)\cosh\left(\frac{\theta_0+\theta_2}{2}\right)-\sinh\left(\frac{\theta_0+\theta_1}{2}\right)\sinh\left(\frac{\theta_0+\theta_2}{2}\right)=\cosh\left(\frac{1}{2}(\theta_1-\theta_2)\right)$.
So $\theta=\frac{1}{2}(\theta_1-\theta_2)$ is independent of $P_0$.
Relation to circular inscribed theorem
This proof is very similar to a proof of the Euclidean inscribed angle theorem for a circle - the hyperbolic trig identity can be replaced by an ordinary trig identity.
Question
Is this result well-known? Wikipedia mentions a similar result for hyperbolae of the form $y=\frac{1}{x}$, but there the 'angle' measure (ratio of gradients of lines) is not obviously related to the Minkowski metric, nor obviously generalisable to other rectangular hyperbolae.
Nonetheless, it seems like this is a result which should be known already. If so, does anybody have a reference?