Hm, I solved this problem with nothing but angle-chasing Dinner now, solution later.

Wow! Very long dinner!

Let $ABCD$ be the trapezoid with $AB\|CD$, and let $O$ be the intersection of the diagonals $AC,BD$. For each point $X\in\{A,B,C,D\}$, let $X'$ be the point we obtain through the described procedure. $A'C'$ is the reflection of $AC$ in $BD$, so it passes through $O$, and $B'D'$ is the reflection of $BD$ in $AC$, and it also passes through $O$. We also have $\frac{OA'}{OC'}=\frac{OA}{OC},\ \frac{OB'}{OD'}=\frac{OB}{OD},\ \frac{OA}{OC}=\frac{OB}{OD}$, so from these we get $\frac{OA'}{OC'}=\frac{OB'}{OD'}$ (and $A'C'$, on the one hand, and $B',D'$ on the other hand are on opposite sides of $O$), so $A'B'\|C'D'$.

Myth wrote: Wow! Very long dinner! Lol, I apologize. Anyway, here we go. In trapezoid ABCD, AB||CD. Let A₁ be the projection of A onto BD, and A₂ be the reflection of A through BD. Define similar points for B, C, and D. We have <BAA₁ = 90 - <ABD = 90 - <CDB = <DCC₁, and similar angle chasing gives <ABB₁ = <CDD₁. Since the quadrilaterals ABB₁A₁ and CDD₁C₁ are also cyclic (because of the right angles), the other two pairs of corresponding angles are also equal, meaning these two quadrilaterals are similar. And since AB||CD, it follows that the corresponding sides of these trapezoids are parallel. Now we chase angles to find that <A₂A₁B₁ = 180 - AA₁B₁ = 180 - CC₁D₁ = <C₂C₁D₁, and similarly <B₂B₁A₁ = D₂D₁C₁. Since we also have C₁C₂/A₁A₂ (*) = CC₁/AA₁ =C₁D₁/A₁B₁ (*) =DD₁/BB₁ =D₁D₂/B₁B₂ (*) (the starred equations are the important ones), the trapezoids A₁A₂B₂B₁ and C₁C₂D₂D₁ have three sides in proportion and the two included angles equal, i.e. they are similar. And since C₁D₁ || A₁B₁, these two trapezoids have parallel corresponding sides. In particular, A₂B₂ || C₂D₂.