$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?

If you know a bit about complex numbers and roots of unity, here is another approach:
The $n^\text{th}$ roots of unity are vertices of a regular $n$-gon. Let $z=1$ denote $A_1$ and $w = e^{\frac{2\pi\iota}{n}}$ denote $A_2$ in the complex plane. Then $w^2$ denotes $A_3$ and $w^3$ denotes $A_4$. Now, the given equation can be written as
$$\frac{1}{|1-w|} = \frac{1}{|1-w^2|} + \frac{1}{|1-w^3|}$$
Multiplying with $|1-w|$ throughout, we get
$$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+w^2|}$$
Now since $|w|=1$, pulling out $|w|$ from the denominator of the second term in the R.H.S, 
$$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+\bar{w}|}$$
Notice that $w+\bar{w}$ is purely real and can be written $w+\frac{1}{w}$, hence
$$1 - \frac{w}{1+w+w^2} = \frac{1+w^2}{1+w+w^2} = \frac{1}{|1+w|}$$
Squaring both sides and using $|z|^2 = z\bar{z}$,
$$\Bigg(\frac{1+w^2}{1+w+w^2}\Bigg)^2 = \frac{1}{(1+w)} \cdot \frac{1}{(1+\frac{1}{w})}= \frac{w}{(1+w)^2}$$
Cross multiply,
$$(1+w+w^2+w^3)^2 = w (1+w+w^2)^2$$
and simplify to get
$$1+w+w^2+w^3+w^4+w^5+w^6 = 0$$
Therefore,
$$1-w^7 =0$$
and $w \neq 1$, so $w$ must be a $7$th root of unity. Argue that $n =7$.
