Are there two non-congruent quadrilaterals with same sets of sides and angles?
It is relatively easy to construct such pentagons. Trying to construct quadrilaterals allows for seemingly multiple degrees of freedom, yet my every attempt ends up not working out.
On
Yes. There are two degrees of freedom in the solution space, so we can add two extra constraints to make the computation easier. Here’s a solution with a 90° angle and a 45° angle:
The side lengths, from longest to shortest, are $$\begin{align*} 1&, \\ α &\approx 0.838045, \\ (2 + \sqrt2) + (6 + 4\sqrt2) α + \sqrt2 α^2 - (12 + 8\sqrt2) α^3 &\approx 0.454531, \\ (3 + 3\sqrt2) + (9 + 7\sqrt2) α - (2 + 2\sqrt2) α^2 - (16 + 12\sqrt2) α^3 &\approx 0.284498, \end{align*}$$
where $α$ is the positive root of
$$1 - \sqrt2 - 2α - 2α^2 + (4 - \sqrt2)α^3 + 4α^4 = 0.$$
No, two non-congruent quadrilaterals with same sets of sides and angles don't exist. Let's prove this:
Suppose that there exist such two quadrilaterals $ABCD$ and $A^{'}B^{'}C^{'}D^{'} $.
We have $AB=A^{'}B^{'}, AD=A^{'}D^{'}$ and angles $A$ and $A^{'}$ are congruent, therefore $\triangle ABD \equiv \triangle A^{'}B^{'}D^{'}$, therefore $BD = B^{'}D^{'}.$
Analogously, we have $AC=A^{'}C^{'}$
Therefore, both quadrilaterals have same length of sides, diagonals, and same angles between any two un-opposed sides and, because of the earlier-proven congruences, same angles between sides and diagonals. Therefore, the two quadrilaterals are congruent, which is contradiction. Q.E.D