If all the sides of an n sided polygon are equal. Is it always a regular polygon?

Question

For an $n$-sided polygon, if all the sides are equal, is it a regular polygon?
If yes then why is it defined to have equal angles?
If not so, how to prove that all angles are equal?

edit: I meant it for $n>4$

edit 2: I tried to play with some figures

Answer 1

The equiangular part of the definition is not a redundancy, if that is what you mean. As a counterexample consider the polygon that appears on the Swiss flag. It is an equilateral dodecagon, 12 sides, all equal in length. It is not equiangular. Eight of the interior angles are 90°, and the other four are 270°. It has only four rotation symmetries and four reflections. A regular dodecagon has 12 of each.

Answer 2

For polygons with more then 3 sides, it is not always a regular polygon. Assume a rhombus, which is a type of parallelogram that has the added restriction of being equilateral. It could have angles that don't all measure 90° (eg two 45° angles and two 135° angles).

That's the case for $n=4$, but what about the rest? For $n>4$ a more general method of generating irregular equilateral polygons would be to have one angle be 90°, and the rest will have measures according to this formula: $$\frac{90(2n-3)}{n-1}$$ Where n is the number of sides of the polygon.