There is a beautiful fact:
If you take a regular N-sided polygon, which is inscribed in the unit circle and find the product of all its diagonals (including two sides) carried out from one corner you will get N exactly:
$A_1A_2\cdot A_1A_3\cdot …\cdot A_1A_N = N$
For example, for a square we have $\sqrt{2}\cdot 2\cdot \sqrt{2} = 4$.
I know there is some prove, which is based on complex numbers. But the result is so simple that I wonder is there much more simple prove, which you can explain to a school boy easily?
P.S. Please, use spoiler tag >! in your answers.
Best Answer
Here is an extremely simple proof but it requires working in the field $\mathbb{C}$.
For any natural $n > 0$:
Let $w = e^{i\frac{2\pi}{n}}$
$\prod_{k=1}^n (z-w^k) = z^n-1$ because both are monic polynomials of degree $n$ with the same $n$ roots
Thus $\prod_{k=1}^{n-1} (z-w^k) = \lim_{t \to z} \frac{t^n-1}{t-1} = \sum_{k=0}^{n-1} z^k$
In particular $\prod_{k=1}^{n-1} (1-w^k) = \sum_{k=0}^{n-1} 1^k = n$
Therefore the product of the diagonals desired is $|n| = n$