Geometrically proving the half-angle formula for sine from a particular diagram 
I have attached a picture of the diagram I am using to prove the trig identity $\sin(\frac{\alpha}{2})= \sqrt{(\frac{1-\cos\alpha}{2})}$. I have that $\sin \alpha = \frac{DG}{OD}$ and $\sin\frac{\alpha}{2}=\frac{DE}{OD}$ as well as $\sin\frac{\alpha}{2}=\frac{EF}{OE}=EF$. I'm unsure of what segment could even represent $\sqrt{\frac{1-\cos\alpha}{2})}$?

Construct $DF$
let $DF$ intersect $OE$ at $P$
$\triangle OPF \cong \triangle OPD$
$\angle OPD\cong \angle OPF$ is a right angle.
$\sin \frac a2 = \frac {\text {opposite}}{\text{hypotenuse}} = \frac {PF}{OF} = \frac {PD}{OD}$
$OF = OD = \cos \frac a2\\
DP = PF = \cos \frac a2\sin \frac a2$
$DF= DP + PF =  2\cos \frac a2\sin \frac a2$
by the Pythagorean theorem $DF^2 = FG^2 + GD^2$
$OF = \cos \frac a2\\OG = OD\sin a= \cos \frac a2 \cos a\\FG = OF - OG = \cos a2 (1-cos a)\\GD = OD\sin a = \cos \frac a2\sin a$
$(2\cos \frac a2\sin \frac a2)^2 = (\cos \frac a2)^2(1-\cos a)^2 + ((\cos \frac a2)^2(\sin a)^2$
We can divide through by $(\cos \frac a2)^2$
$(2\sin \frac a2)^2 = (1-\cos a)^2 + (\sin a)^2$
$4\sin^2 \frac a2 = 1-2\cos a + cos^2 a + \sin^2 a = 2-2\cos a\\
\sin \frac a2 =  \sqrt {\frac{1-\cos a}{2}}$
