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?
Fact#1 $AB = BC = s$, say
Fact#2 $\angle C = \pi – \angle A$.
Fact#3 $\cos \angle C = cos (\pi – \angle A) = – cos \angle A$
Let $MA = a, MB = b, MC = c$.
By cosine law, $b^2 = s^2 + a^2 – 2sa \cos \angle A$
$∴ b^2 = s^2 + a^2 + 2sa \cos \angle C$ --------(1)
Also, $b^2 = s^2 + c^2 – 2sc \cos \angle C$ ------(2)
Result follows by applying $(1)*bc + (2)*ab$ (after simplification).