In an isosceles trapezoid, $ABCD$ ($AB ∥ CD$) points $E$ and $F$ lie on the segment $CD$ in
such a way that $D$, $E$, $F$, and $C$ are in that order, and $DE = CF$.
Let $X$ and $Y$ be the reflection
of $E$ and $C$ with respect to $AD$ and $AF$.
Prove that circumcircles of triangles $\triangle ADF$ and $\triangle BXY$
are concentric.
If $O$ is the circumcentre then I am trying to prove that $\triangle ODX$ is congruent to $\triangle OFY$. Now we already know that $OD = OF$ and $DX = FY$. Now I just want to prove that angle $O\hat{D}X = O\hat{F}Y$. With this I'll at least be able to prove the centre is equidistant from two vertices and then try to infer something for the third vertice.
Also, I'm sorry if my question is a duplicate but the similar questions feature just doesn't work out for me... If you could please include the way you thought of the solution along with it, it would mean the world to me. Appreciate it!
Your guess is actually working. Let's assume that $O$ is the circumcircle of $\triangle ADF$. Note that:
$$\angle OFY=180^{\circ}-\angle DFO-2\angle AFD, \\ \angle XDO=\angle FDO+2\angle ADF.$$
Having $\angle DFO=\angle FDO= \angle DAF-90^{\circ}$, we easily get: $\angle OFY=\angle XDO.$ Henece, as you've already mentioned, $|XO|=|YO|$.
Moreover, note that $|AO|=|OF|$ and $|BF|=|AE|=|AX|$, and:
$$\angle XAO=2\angle DAE +\angle EAO=\angle DAE+\angle DAO=\angle DAE+\angle ADO\\= \angle DAE+\angle ADF+\angle FDO \\=\angle CBF+\angle BCF+\angle DFO\\=\angle BFD+\angle DFO=\angle BFO.$$
Therefore, $\triangle XAO$ and $\triangle OFB$ are congruent. So, $|XO|=|OB|.$
We are done.