If $|z_1|=|z_2|=|z_3|$
I have used:
$z_1=x_1+\mathrm iy_1,z_2=x_2+\mathrm iy_2,z_3=x_3+\mathrm iy_3$
and obtained $$\arg\frac{z_3-z_2}{z_3-z_1} = \arctan \frac{(y_3-y_2)(x_3-x_1)-(y_3-y_1)(x_3-x_2)}{(x_3-x_2)(x_3-x_1)}$$
But, then I'm stuck.
Urgent help needed.
Best Answer
$arg({a \over b})=arg(a)-arg(b)$
So your problem becomes...
$arg(z_3-z_2)-arg(z_3-z_1)=\frac{1}{2}(arg(z_2)-arg(z_1))$
Now let $z_k=re^{i\theta_k}=r(\cos{\theta_k+i\sin{\theta_k}})$
$arg(r(\cos{\theta_3}-\cos{\theta_2}+i(\sin{\theta_3}-\sin{\theta_2})))-arg(r(\cos{\theta_3}-\cos{\theta_1}+i(\sin{\theta_3}-\sin{\theta_1})))=\frac{1}{2}(arg(r(\cos{\theta_2+i\sin{\theta_2}}))-arg(r(\cos{\theta_1+i\sin{\theta_1}})))$
$arctan({\sin{\theta_3}-\sin{\theta_2} \over \cos{\theta_3}-\cos{\theta_2}})-arctan({\sin{\theta_3}-\sin{\theta_1} \over \cos{\theta_3}-\cos{\theta_1}})={1 \over 2}(arctan({\sin{\theta_2} \over \cos{\theta_2}})-arctan({{\sin{\theta_1} \over \cos{\theta_1}}}))$
It can be proven with trig identities that ${{\sin{x}-\sin{y}} \over {\cos{x}-\cos{y}} }={\cot({\frac{x+y}{-2}})}$
$arctan({\cot({\frac{\theta_3+\theta_2}{-2}})})-arctan({\cot({\frac{\theta_3+\theta_1}{-2}})})={1 \over 2}(arctan(\tan{\theta_2})-arctan(\tan{\theta_1})$
It can be shown that $arctan(\cot{x})={\pi \over 2}-x$
$({\pi \over 2} - \frac{\theta_3+\theta_2}{-2})-({\pi \over 2} - \frac{\theta_3+\theta_1}{-2})={1 \over 2}({\theta_2}-{\theta_1})$
$\frac{\theta_3+\theta_2}{2}-\frac{\theta_3+\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$
$\frac{\theta_3+\theta_2-\theta_3-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$
$\frac{\theta_2-\theta_1}{2}={1 \over 2}({\theta_2}-{\theta_1})$
The statement appears to be proven, but be aware that this makes many assumptions, largest of which is that the angles all exist in the first quadrant. If the angles were all in different quadrants, the proof would be much more involved, and to make such a proof would need a lot of meticulous calculations and studying many different cases - mostly because of the arctan function and its limited range.
Good luck.