Complicated Planar Geometry A regular polygon $\mathcal{P}$ is inscribed in a circle $\Gamma$. Let $A$, $B$, and $C$ be three consecutive vertices of the polygon $\mathcal{P}$, and let $M$ be a point on the arc $AC$ of $\Gamma$ that does not contain $B$. Prove that
$MA \cdot MC = MB^2 - AB^2.$
I have tried inscribing the polygon in a unit circle with no luck.  Any help is greatly appreciated!
 A: take the diameter of the circumcircle to be $2.$ let $$\angle MAC = \angle MBC = \alpha, \angle MBA = \angle MCA = \beta, \angle BMC = \angle BMA = \angle BAC = \angle ABC = \gamma.$$ observe that $$\alpha + \beta + 2\gamma = 180^\circ$$
then by the rule of $sine,$ we have $$MC = \sin \alpha, MA = \sin \beta, MB = \sin(\beta + \gamma), AB = \sin \gamma.$$
now, $$\begin{align}MB^2 - AB^2 &= \sin^2(\beta+\gamma) - \sin^2\gamma \\
&=(\sin(\beta+\gamma)-\sin\gamma)(\sin(\beta+\gamma)+\sin \gamma)\\
&=2\cos(\gamma + \beta/2)\sin (\beta/2)2\sin(\gamma + \beta/2)\cos (\beta/2)\\
&=\sin\beta\sin(2\gamma + \beta)\\
&=\sin\beta\sin \alpha\\
&=MA \cdot MC \end{align}$$
A: Let $M=e^{ix}$, $B=1$, $A=e^{ia}$, $C=e^{-ia}$, so
$MA=|e^{ix}-e^{ia}|, MC=|e^{ix}-e^{-ia}|$, 
$$\begin{align}MA\cdot MC = &|(e^{ix}-e^{ia})(e^{ix}-e^{-ia})| =\\ 
&|e^{2ix}-e^{ix}(e^{ia}+e^{-ia})+1| =\\
&|e^{2ix}-2e^{ix}+1-e^{ix}(e^{ia}-2+e^{-ia})| =\\
&|(e^{ix}-1)^2-e^{ix}(e^{i\frac{a}{2}}-e^{-i\frac{a}{2}})^2|\hbox{, where}\end{align}$$
$|(e^{ix}-1)|=MB, (e^{i\frac{a}{2}}-e^{-i\frac{a}{2}})=i\cdot AB$,
we only need to show that $arg((e^{ix}-1)^2)=-(arg(-e^{ix})+\frac{\pi}{2})$, that is "an inscribed angle equals a half of its corresponding central angle".
So we would have $$\begin{align}|(e^{ix}-1)^2-e^{ix}(e^{i\frac{a}{2}}-e^{-i\frac{a}{2}})^2|=\\
|(e^{ix}-1)^2|-|e^{ix}(e^{i\frac{a}{2}}-e^{-i\frac{a}{2}})^2|=\\
|(e^{ix}-1)^2|-|e^{ix}|\cdot|(e^{i\frac{a}{2}}-e^{-i\frac{a}{2}})^2|=\\MB^2-1\cdot AB^2\end{align}$$
