4-gon with two opposite sides congruent and two opposite angles congruent: necessarily a parallelogram? [duplicate]

My kid's geometry classmate tried to use the following "theorem" in a proof:

If a quadrilateral has a pair of opposite sides congruent and a pair of opposite angles congruent, it's a parallelogram.

The teacher (correctly) didn't allow this, as they hadn't proven the "theorem" in class. The teacher, moreover, conjectured but could not prove that the "theorem" was in fact false. Is it?

Consider isosceles trapezoid $$ABCD$$ with $$\overline{AB}\Vert\overline{CD}$$ and $$AD=BC$$. Cut it into two triangles along diagonal $$\overline{AC}$$ and glue them together by identifying $$\overline{AD}$$ to $$\overline{BC}$$, $$A$$ to $$B$$, $$D$$ to $$C$$. The resulting quadrilateral has a pair of opposite sides congruent because they were the diagonal $$\overline{AC}$$ of the trapezoid, and a pair of opposite angles congruent because they were corresponding angles $$(\angle BAC,\angle ACD)$$ across a transversal of parallel lines.
Moreover, every quadrilateral (without intersecting sides) that's not a parallelogram but has a pair of opposite sides congruent and a pair of opposite angles congruent can be constructed using that method. Indeed, consider such a quadrilateral $$MNOP$$, WLOG $$MN=OP$$ and $$\angle N\cong\angle P$$. Cut it along $$\overline{MO}$$ and glue the resulting triangles by identifying $$\overline{MN}$$ to $$\overline{PO}$$, $$M$$ to $$P$$, $$N$$ to $$O$$. The resulting quadrilateral has two congruent sides from $$\overline{MO}$$ and the other two sides are parallel because corresponding angles $$(\angle N,\angle P)$$ across the transversal are congruent. Moreover, it's not a parallelogram: if it were, the other pair of opposite sides would be congruent, $$NO=MP$$, so the original $$MNOP$$ would be a parallelogram also. Thus it's an isosceles trapezoid, as required.