Prove that
$\arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \frac{1}{2} \arg\left(\frac{z_{2}}{z_{1}}\right)$
if $|z_{1}|=|z_{2}|=|z_{3}|$.
Prove that
$\arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \frac{1}{2} \arg\left(\frac{z_{2}}{z_{1}}\right)$
if $|z_{1}|=|z_{2}|=|z_{3}|$.
Note that $\arg{w_2-w_0\over w1-w_0}=\angle(w_1,w_0,w_2)$. With the help of a figure you can easily verify that the stated formula is an immediate consequence of the theorem about central and peripheral angles.
This is simply an application of the Inscribed Angle Theorem. Since $ \frac{z_3-z_2}{z_3-z_1}=\frac{z_2-z_3}{z_1-z_3} $ and the Inscribed Angle Theorem says that $ 2\,\arg\left(\frac{z_2-z_3}{z_1-z_3}\right)=\arg\left(\frac{z_2}{z_1}\right)\pmod{2\pi} $
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we get that $ 2\,\arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\arg\left(\frac{z_2}{z_1}\right)\pmod{2\pi} $ which can be rewritten as $ \arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\frac12\arg\left(\frac{z_2}{z_1}\right)\pmod{\pi} $
This is not exactly the same and points out that the relation is not always true.
For example, suppose $z_1=\frac{1-i}{\sqrt{2}}$, $z_2=\frac{1+i}{\sqrt{2}}$, and $z_3=1$. Then $ \arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\frac54\pi $ yet $ \frac12\arg\left(\frac{z_2}{z_1}\right)=\frac\pi4 $
Hint. Assume $|z| = 1$.
$\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} = (\cos \alpha + i \sin \alpha)(\cos \beta - i \sin \beta) = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$
$\cos \alpha + i\sin \alpha - \cos \beta - i\sin \beta = -2 \sin(\frac{\alpha - \beta}{2})\sin(\frac{\alpha + \beta}{2}) + 2i\sin(\frac{\alpha - \beta}{2})\cos(\frac{\alpha + \beta}{2}) =\\ 2\sin(\frac{\alpha - \beta}{2})\Big(\cos(\frac{\pi}{2}+\frac{\alpha + \beta}{2}) + i\sin(\frac{\pi}{2}+\frac{\alpha + \beta}{2}) \Big)$
Deduce how you can express your equation in terms of $\arg z_1$, $\arg z_2$ and $\arg z_3$.
Let $z_j=R(\cos2t_j+i\sin2t_j)$ where $j=1,2,3$ and $R \neq 0$
So, $\frac{z_3-z_2}{z_3-z_1}=\frac{R(\cos2t_3+i\sin2t_3)-R(\cos2t_2+i\sin2t_2)}{R(\cos2t_3+i\sin2t_3)-R(\cos2t_1+i\sin2t_1)}$
$=\frac{(\cos2t_3-\cos2t_2)+i(\sin2t_3-\sin2t_2)}{(\cos2t_3-\cos2t_1)+i(\sin2t_3-\sin2t_1)}$
$=\frac{-2\sin(t_3-t_2)\sin(t_3+t_2)+2i\sin(t_3-t_2)\cos(t_3+t_2)}{-2\sin(t_3-t_1)\sin(t_3+t_1)+2i\sin(t_3-t_1)\cos(t_3+t_1)}$ applying $\sin C-\sin D$ and $\cos C-\cos D$
$=\frac{2i\sin(t_3-t_2)(\cos(t_3+t_2)+i\sin(t_3+t_2))}{2i\sin(t_3-t_1)(\cos(t_3+t_1)+i\sin(t_3+t_1))}$
$=\frac{\sin(t_3-t_2)e^{i(t_3+t_2)}}{\sin(t_3-t_1)e^{i(t_3+t_1)}}$ as $e^{ix}=\cos x+i\sin x$
$=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}e^{i(t_2-t_1)}$
$=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}(\cos(t_2-t_1)+i\sin(t_2-t_1))=X+iY(say)$
So, $X=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}\cos(t_2-t_1)$ and $Y=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}\sin(t_2-t_1)$
So, $\frac Y X = \tan (t_2-t_1)$ assuming $\pi ∤ (t_3 -t_1)$
$\implies \arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \arctan \left(\frac Y X\right) $ $= t_2-t_1=\frac{1}{2}(2t_2-2t_1)=\frac{1}{2}(\arg z_2 -\arg z_1)=\frac{1}{2}\arg{\frac{z_2}{z_1}}$