Do equal angles necessarily mean a polygon is regular? In a polygon, if all the sides are equal, it doesn’t necessarily mean that the polygon is regular (eg. a rhombus). Is this also true for angles? Meaning can you draw a polygon whose interior angles are equal, but the shape is still not regular? I couldn’t think of any examples, but I’m sure there is one.
 A: Here are four pentagons all with interior angles of $108^\circ$. Only the largest is regular. The generalization to any regular polygon should be clear.

A: Start with any polygon that has more than three edges. Move one of the edges parallel to itself a little, extend or contract the adjacent edges appropriately and you will have a new polygon with the same edge directions but different relative side lengths. If you start with a regular polygon the angles will remain all the same.

The idea behind this construction is generic. If you start with any sequence of $n > 3$ vectors that span the plane there will be an $n-2$ dimensional space of linear combinations that vanish. Each such linear combination defines a polygon with the same edge directions: form the partial sums in order to find the vertices.
A: The side lengths need not be equal. It is amenable to a vector representation. 
If  segments of length of an isotropic force set $F_i$  acts on a particle ( components ${F_{xi},F_{yi}}$ of consecutive positive integers $i$ )  then the vectors sum up to zero: 
$$ \Sigma_i^n F_{x}=0 $$
$$ \Sigma_i^n F_{y}=0 $$
Each vector joins tail to head of vector and the force polygon comes  back to point of start after all $n$ rotations of $\dfrac{2 \pi}{n}$.
So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.
