geometry
Is this phrase safe to consider in general:
Equal sides of a polygon have corresponding equal angles
if not how would you refine or correct it
Example of a corresponding angle would be
Edited:
For example suppose you ignore the fact that the above triangle is an equilateral and treat it as a regular polygon.Then according to the above statement Angle A and Angle B are both equal because Side A corresponds to angle A which is equal to SideB which corresponds to angle B
Consider the following triangle $\triangle ABC$, the $ \color{maroon} {\text{poly 1}}$ below, with sides $\color{maroon}{\overline{AB}=c}$ and $\color{maroon}{\overline{AC}=b}$ known. Further the angle between them, $\color{green}\alpha$ is known.
$\hskip{2 in}$
Then, the law of cosines tell you that $$\color{maroon}{a^2=b^2+c^2-2bc\;\cos }\color{blue}{\alpha}$$
You can use tetrahedral Laws of Cosines to get these angles.
Let $W$, $X$, $Y$, $Z$ be the areas of faces opposite respective vertices $O$, $A$, $B$, $C$. Write "$\angle PQ$" for the dihedral angle along edge $\overline{PQ}$. Then we have what I call the "First" Law of Cosines: $$\begin{align}W^2 &= X^2 + Y^2 + Z^2 - 2 Y Z \cos \angle OA - 2 Z X \cos \angle OB - 2 X Y \cos \angle OC & (1)\end{align}$$ as well as this "Second" one: $$\begin{align} Y^2 + Z^2 - 2 Y Z \cos \angle OA &= W^2 + X^2 - 2 W X \cos \angle BC \\[4pt] Z^2 + X^2 - 2 Z X \cos \angle OB &= W^2 + Y^2 - 2 W Y \cos \angle CA &(2) \\[4pt] X^2 + Y^2 - 2 X Y \cos \angle OC &= W^2 + Z^2 - 2 W Z \cos \angle AB \end{align}$$
In a tri-rectangular (or "right-corner") tetrahedron, with mutually-orthogonal edges $\overline{OA}$, $\overline{OB}$, $\overline{OC}$, the corresponding dihedral angles are right angles whose cosines vanish. The First Law of Cosines reduces to the tetrahedral Pythagorean Theorem:
$$W^2 = X^2 + Y^2 + Z^2 \qquad (1^\star)$$
while the aspects of the Second Law simplify to $$X = W \cos \angle BC \qquad Y = W \cos \angle CA \qquad Z = W \cos \angle AB \qquad (2^\star)$$
Faces $X$, $Y$, $Z$ being right triangles, the individual areas are given by $$X = \frac{1}{2}|\overline{OB}||\overline{OC}| \qquad Y = \frac{1}{2}|\overline{OC}||\overline{OA}| \qquad Z = \frac{1}{2}|\overline{OA}||\overline{OB}|$$ with $W$ computed via $(1^\star)$.
Another approach to $(2^\star)$ ...
If you look at the tetrahedron with, say, edge $\overline{BC}$ directed at your eye, then $\overline{BC}$ collapses to a point (call it $D$) and the tetrahedron collapses to a right triangle $\triangle OAD$ with right angle at $O$. Leg $\overline{OD}$ and hypotenuse $\overline{AD}$ are projections of altitudes from $O$ and $A$ dropped to $\overline{BC}$ in the tetrahedron. Since $$X = \frac{1}{2} |\overline{BC}| |\overline{OD}| \qquad W = \frac{1}{2} |\overline{BC}||\overline{AD}|$$ we have $$\cos \angle BC = \cos D = \frac{|\overline{OD}|}{|\overline{AD}|} = \frac{2X/|\overline{BC}|}{2W/|\overline{BC}|} = \frac{X}{W}$$
Edit. The updated question asks specifically about a right-corner tetrahedron $OABC$ where $|\overline{OA}| = |\overline{OB}| = |\overline{OC}|$. Writing "$a$" for that common length, we have $$X = Y = Z = \frac{1}{2}a^2 \qquad \qquad W^2 = 3 X^2 = \frac{3}{4}a^4 \quad\to\quad W = \frac{\sqrt{3}}{2} a^2$$ so that $$\cos \angle BC = \cos \angle CA = \cos \angle AB = \frac{X}{W} = \frac{\frac{1}{2}a^2}{\frac{\sqrt{3}}{2}a^2} = \frac{1}{\sqrt{3}}$$ whence $$\angle BC = \angle CA = \angle AB = \operatorname{acos}\frac{1}{\sqrt{3}} \approx 54.7356^\circ$$
Best Answer
Construct a pentagon of all equal sides. Physically, out of pieces of straws or something. Note how you can bend it in any which way so that no angles are the same. So the question of "equal sides correspond to equal angles" is true for triangles, can maybe be rephrased in a way that's true for quadrilaterals (since there it forces two pairs of equal angles), but definitely fails to work in general.
A better statement to prove might be the reverse: Do equal angles of a polygon imply equal sides?