A Polygon is inscribed in a circle $\Gamma$ A regular polygon P is inscribed in a circle $\Gamma$. Let A, B, and C, be three consecutive vertices on the polygon 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 inscribed the polygon P in the unit circle and let B=1. Past this, I'm not really sure how to proceed. It might be helpful to have M at 1 as well but given that the polygon would have the n-th roots of unity in it B seemed like a good choice. What would be a good way to proceed from here?

We can place the diagram in the complex plane so that $\Gamma$ is the unit circle, point $A$ goes to the complex number $a$, point $B$ goes to the complex number 1, and point $C$ goes to the complex number $\overline{a} = 1/a$. Let $m$ be the complex number corresponding to point $M$.
Then
\begin{align*}
MA \cdot MC &= |m - a| \cdot \left| m - \frac{1}{a} \right| \\
&= |a - m| \cdot \left| \frac{am - 1}{a} \right| \\
&= |a - m| \cdot |am - 1| \\
&= |(a - m)(am - 1)|.
\end{align*}
Also,
\begin{align*}
MB^2 - AB^2 &= |m - 1|^2 - |a - 1|^2 \\
&= (m - 1)(\overline{m} - 1) - (a - 1)(\overline{a} - 1) \\
&= (m - 1) \left( \frac{1}{m} - 1 \right) - (a - 1) \left( \frac{1}{a} - 1 \right) \\
&= 1 - m - \frac{1}{m} + 1 - 1 + a + \frac{1}{a} - 1 \\
&= \frac{a^2 m - am^2 - a + m}{am} \\
&= \frac{(a - m)(am - 1)}{am}.
\end{align*}
Since $M$ lies on the arc $AC$ that does not contain $B$, $MB^2 - AB^2$ is positive, so we can take the absolute value of this expression, to get
\begin{align*}
MB^2 - AB^2 &= \left| \frac{(a - m)(am - 1)}{am} \right| \\
&= \frac{|(a - m)(am - 1)|}{|am|} \\
&= |(a - m)(am - 1)|.
\end{align*}
Hence, $MA \cdot MC = MB^2 - AB^2$.
