I let AB = 1, and letting AC = x, and AD = y set up
1 = $\frac{1}{x}$ + $\frac{1}{y}$ , multiplied to get xy = y + x, added 1, and got (x-1)(y-1) = 1
I know AC and AD are equal, but don't know how to get the number of sides of the polygon with this information. Any ideas? Any and all help is appreciated!
Let $n \ge 4$ be the number of vertices of the regular polygon. I work in the complex plane, so I may assume that $A, B, C, D$ are represented by the complex numbers $re^{ik2\pi/n}$ for $k=0,1,2,3$.
So we have to solve $|e^{i2\pi/n}-1|^{-1} = |e^{i4\pi/n}-1|^{-1}| + |e^{i6\pi/n}-1|^{-1}$, namely (using Euler formulas) $\sin(\pi/n)^{-1} = \sin(2\pi/n)^{-1}+\sin(3\pi/n)^{-1}$.
Multiplying both sides by $\sin(\pi/n)$ and using trigonometric formulas yields $$1 = \frac{1}{2\cos(\pi/n)} + \frac{1}{4\cos^2(\pi/n)-1},$$ $$2\cos(\pi/n)(4\cos^2(\pi/n)-1) = 4\cos^2(\pi/n)-1 + 2\cos(\pi/n).$$ Hence $\cos(\pi/n)$ is a solution of the equation $8x^3-4x^2-4x+1=0$.
On the other hand, set $a = \cos(2\pi/7)$, $b = \cos(4\pi/7)$, $c=\cos(6\pi/7)$. Then $a>b>c$ since $\cos$ decreases on $[0,\pi]$. $$1 + 2a + 2b + 2c = \sum_{k=-3}^3 e^{i2k\pi/7} = 0.$$ Trigonometric identities yields $b=2a^2-1$ and $c=4a^3-3a$. Hence $8a^3+4a^2-4a-1=0$.
We also have $8b^3+4b^2-4b-1=0$ since $c = \cos(8\pi/7) = 2b^2-1$ and $a = \cos(12\pi/7) = 4b^3-3b$.
In the same way, $8c^3+4c^2-4c-1=0$ since $a = \cos(12\pi/7) = 2c^2-1$ and $b = \cos(18 \pi/7) = 4c^3-3c$.
As a result, the three roots of the equation $8x^3-4x^2-4x+1=0$ are $-a = \cos(5\pi/7)$, $-b = \cos(3\pi/7)$, $-c = \cos(\pi/7)$.
Since $\cos$ decreases on $[0,\pi]$, we derive $\pi/n = \pi/7$, $3\pi/7$ or $5\pi/7$. The only possibility is $n=7$.
