Equilateral pentagon with increasing/equal angles 
Suppose all sides of a convex pentagon ABCDE have the same size, and $\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E $. Prove that this pentagon is a regular pentagon.

I know this should be proved using contradiction, but I'm not sure how to reach that point. Can anyone guide me to the right solution?
 A: WOLOG, we only need to consider the case where the side of the equilateral pentagon is $1$ and the vertices $A,B,C,D,E$ are ordered counterclockiwisely
on circumference of the pentagon.
Let $\alpha,\beta,\gamma,\delta,\epsilon$ be the external angles at $A, B, C, D, E$. We have
$$\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E
\quad\implies\quad \alpha \le \beta \le \gamma \le \delta \le \epsilon
$$
Since $2\pi = \alpha + \beta + \gamma + \delta + \epsilon \le 5 \epsilon$,
we have $\epsilon \ge \frac{2\pi}{5}$. This leads to
$$\alpha + \beta + \gamma + \delta \le \frac{8\pi}{5}
\implies \alpha + \beta \le \frac{4\pi}{5}
\implies \alpha \le \frac{2\pi}{5}
$$
Choose a coordinate system so that $E = (-\frac12,0), A = (\frac12,0)$.
In this coordinate system, it is not hard to see the $x$-coordinate of $C$ is
$$\frac12 + \cos\alpha + \cos(\alpha+\beta)
\ge \frac12 + \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = 0\tag{*1}$$
THis means $C$ is lying in the left-half plane and hence $|AC| \le |EC|$.
However 
$$\begin{align} &|AC|^2 = |AB|^2 + |BC|^2  + 2|AB||BC|\cos\beta = 2(1+\cos\beta)\\
& |CE|^2 = |CD|^2 + |DE|^2 + 2|CD||DE|\cos\delta = 2(1+\cos\delta)
\end{align}
$$
Together with $\beta \le \delta$, we obtain $|AC| \ge |EC|$. 
Combine with above, we get $|AC| = |EC|$. This forces $C$ to lies on the $y$-axis. 
Notice if either $\alpha < \frac{2\pi}{5}$ or $\alpha + \beta < \frac{4\pi}{5}$, the inequality in $(*1)$ becomes strict. For $C$ to lies on the $y$-axis, we need
$\alpha = \frac{2\pi}{5}$. This leads to
$$2\pi = 5\alpha \le \alpha + \beta + \gamma + \delta + \epsilon = 2\pi$$
Since $\alpha \le \beta \le \gamma \le \delta \le \epsilon$, this forces
$\alpha = \beta = \gamma = \delta = \epsilon$ and
$ABCDE$ is a regular pentagon.
