Finding this complex number relationship using geometry

Show that $|e^{i\theta}-1| = 2|\sin (\frac{\theta}{2})|$ for all $\theta \in \mathbb{R}$ by using the geometry of the triangle with vertices $0,1$ and the midpoint of the line joining $0$ and $e^{i\theta}$.

I'm curious to see how I can use this midpoint to gain this result. This is what I did to get the result:

Drawing up the argand diagram, we can see by the cosine rule that
$$|e^{i\theta} - 1|^2 = 1^2 + |e^{i\theta}|^2 - 2(1^2)(|e^{i\theta}|^2)\cos \theta\\= 2 - 2\cos\theta = 2(2\sin^2(\frac{\theta}{2}))$$
and the result follows by taking the positive root.

Answer

An easier way: Let $A$, $B$ and $C$ be the points $0$, $1$ and $\frac12(1+e^{i\theta})$ (so $C$ is that midpoint). Then the triangle
$ABC$ has a right angle at $C$ and so $|BC|=|AB|\sin\angle CAB
=\sin(|\theta|/2)$. The distance from $1$ to $e^{i|\theta|}$ is twice

$|BC|$.