Complex number identity by trigonometry

Show that $\lvert e^{i\theta} - 1\rvert = 2\lvert\sin(\theta/2)\rvert$ by using the geometry of the triangle with vertices 0, 1, and the midpoint of the line joining 0 and $e^{i\theta}$.

I have been able to show this identity through other means, however I am stuck on how to utilize this particular triangle.

I'm not sure I see what's so important about the suggested triangle, either.

Here's a diagram that illustrates the relation in question:

For the sake of completeness, here's the fairly obvious proof: With $O$, $P$, $Q$, $R$ as shown, we see that $\overline{PQ}$ is both parallel and congruent to $\overline{OR}$, so that $\square ORPQ$ is a parallelogram. This implies $|\overline{OQ}| = |\overline{RP}|$. But $\triangle ORP$ is isosceles, with vertex angle $\theta$ and legs of length $1$, so that (as one readily shows) its base has length $|\overline{RP}| = 2\sin(\theta/2)$, which completes the proof of the relation $|\exp(i\theta)-1| = 2\sin(\theta/2)$ (for appropriately-restricted values of $\theta$; the general case needs a slight tweak). But that's not the point here.

The suggested triangle is $\triangle ORM$, where $M$ is the midpoint of $\overline{OP}$. It's immediately clear that $\triangle PMQ$ is congruent to this triangle, which again sets up the parallelogram-ness of $\square ORPQ$, so that we can argue as above. However, this approach doesn't seem to do justice to the suggestion of "using the geometry of [$\triangle ORM$]" to prove the relation; rather, it uses the triangle as a minor, and soon-ignored, stepping stone on the path to the parallelogram property ... a property that can be proven more directly without referencing the triangle at all.

Maybe, as just indicated by @WillOrrick in a comment to OP, or as inadvertently assumed in a deleted answer, the actual intent of the instruction was to suggest using the geometry of $\triangle ORN$, where $N$ is the midpoint of $\overline{RP}$, aka, the midpoint of $1$ and $\exp(i\theta)$.

That is, perhaps the question had a typo, replacing "$1$" with "$0$" when defining the helpful midpoint.

$|e^{i\theta}-1|=|\cos\theta+i\sin\theta-(1+0i)|=|(\cos\theta-1)+i\sin\theta|=\sqrt{(\cos\theta-1)^2+(\sin\theta)^2}=\sqrt{\cos^2\theta+1-2\cos\theta+\sin^2\theta}=\sqrt{2-2\cos\theta}=\sqrt{2(1-\cos\theta)}=\sqrt{2(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}-\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2})}=\sqrt{4sin^2\frac{\theta}{2}}=2|\sin\frac{\theta}{2}|$