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. 
 A: 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.

A: $|e^{i\theta}-1|=|\cos\theta+i\sin\theta-(1+0i)|=|(\cos\theta-1)+i\sin\theta|=\sqrt{(\cos\theta-1)^2+(\sin\theta)^2}=\sqrt{\cos^2\theta+1-2\cos\theta+\sin^2\theta}=\sqrt{2-2\cos\theta}=\sqrt{2(1-\cos\theta)}=\sqrt{2(\sin^2\frac{\theta}{2}+\cos^2\frac{\theta}{2}-\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2})}=\sqrt{4sin^2\frac{\theta}{2}}=2|\sin\frac{\theta}{2}|$
A: Assuming $\theta$ is real and $\theta/\pi$ is not  an integer, let $\theta=\alpha+n\pi$ where $n$ is an integer and $-\pi<\alpha<\pi$. The line segments from $0$ to $1$, and from $0$ to $exp(i\theta)$, each have length $1$, while the  angle between these segments is $abs(\alpha)$. By elementary geometry, 
the distance $D$ from $0$ to $P$, where $P$ is the midpoint of the line segment from $1$ to $exp(i\theta)$ , is $sin(abs(\alpha/2)=abs(sin(\alpha/2))=abs(sin(\theta/2))$. The distance between any two complex numbers $x$ and $y$ is $abs(x-y)$, so the distance from $1$ to $exp(i\theta)$, which is $2D$, is $2D=abs(exp(i\theta)-1)$. QED.   BTW, the step $sin(abs(\alpha/2)=abs(sin(\alpha/2))$  is valid  because $0<\alpha<\pi$.
A: Here is a demonstration using the triangle you specify 
Let $M$ be the midpoint of $OP$ where $P=e^{i\theta}$ and $Q$ is the point $1$. Let angle $OMQ=\phi$
The sine rule in triangle $OMQ$ gives $$\frac{1}{\sin \phi}=\frac{MQ}{\sin\theta}$$
The sine rule in triangle $MPQ$ gives $$\frac{PQ}{\sin \phi}=\frac{MQ}{\sin(\frac 12(\pi-\theta))}$$
Eliminating $MQ$ gives $$PQ=|e^{i\theta}-1|=\frac{\sin\theta}{\sin(\frac 12(\pi-\theta))}= 2\sin(\frac{\theta}{2})$$
I have not inserted all the relevant modulus signs, but clearly they can be. I accept that this is somewhat artificial, and the identity has already been proved by easier means, but this is just a way of involving the triangle you require.
A: 
In triangle $ABC$, consider $A$ as the the vertex $0$, $B$ as the vertex $1$ and $C$ as the vertex $e^{i\theta}$. So the angle $\widehat {BAC}=\theta$
$D$ is the midpoint of the line joining $0$ and $e^{i\theta}$.
We want to use the geometry of $ABD$.
In $ABD$, draw the median from the vertex $D$ now in $ADE$ we have:  
1-$\vec {ED}=\vec{AD}-\vec{AE}$
2-Cosine law in $ADE$: $|\vec{ED}|^2=|\vec{AD}|^2+|\vec{AE}|^2-2|\vec{AD}||\vec{AE}|\cos\theta$
So: $$|\vec{AD}-\vec{AE}|=|\vec{AD}|^2+|\vec{AE}|^2-2|\vec{AD}||\vec{AE}|\cos\theta$$
We have:
$\vec {AD}=\frac{1}{2}e^{i\theta}\Rightarrow |\vec {AD}|=\frac{1}{2}|e^{i\theta}|=\frac{1}{2}(1)=\frac{1}{2}$
$\vec {AE}=\frac{1}{2}+0i=\frac{1}{2}\Rightarrow |\vec {AE}|=\frac{1}{2}$
So
$$|\frac{1}{2}e^{i\theta}-\frac{1}{2}|^2=(\frac{1}{2})^2+(\frac{1}{2})^2-2(\frac{1}{2})(\frac{1}{2})\cos\theta$$$$(\frac{1}{2}|e^{i\theta}-1|)^2=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}\cos\theta$$$$\frac{1}{4}|e^{i\theta}-1|^2=\frac{1}{2}-\frac{1}{2}\cos\theta$$$$|e^{i\theta}-1|^2=2-2\cos\theta=2(1-\cos\theta)$$$$|e^{i\theta}-1|^2=4\sin^2\frac{\theta}{2}$$$$|e^{i\theta}-1|=2|\sin\frac{\theta}{2}|$$
