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?

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.
