Interesting problem. Construct ${G}$ on line $AC$ such that $DEFG$ is an isoleces trapezoid whose diagonals meet at ${H}$. Let points $A', C'$ on line $AC$ such that $\angle EC'A'=\angle DA'C'=90^{\circ}-\angle EFD$.We only need $A'C'=DE$, which finishes the proof. By $DH=HG$ and $\angle DA'C'=90^{\circ}-\angle EFD=90^{\circ}-\angle HGD$, ${H}$ is the circumcenter of $\triangle {A'DG}$, so $HA'=HG$. SImilarly, $HC'=HF$, so by symmetry $A'C'=FG=DE$. Done!

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Here is a solution using spiral symmetry. Let the intersection of $AC$ and $DE$ be $X$. Notice $XE=XF$. Let $S$ be the center of the unique rotation that maps segment $AC$ to $DE$ which is well known to be the intersection of the circumcircles of $ABC$ and $BDE$. Let $T$ be the image of $B$ under this rotation. Lemma 1: Line $SX$ is the perpendicular bisector of segment $EF$ It is sufficient to show that $SX$ is the angle bisector of $\angle EXF$ but this is trivial due to the fact that $ASC$ and $DSE$ are similar so the length of the altitude from $S$ to $AC$ and $DE$ are equal. Lemma 2: The points $S$, $X$, and $T$ are collinear Let $\Phi$ be the angle of the spiral symmetry described earlier. Let $TS$ and $DE$ intersect at $X'$. Then, $$\angle SX'D=\frac{1}{2}\overarc{DT}-\frac{1}{2}\overarc{ES}=\frac{1}{2}\overarc{ST}=\frac{1}{2}(180-\Phi)=\angle SXD$$Thus $X$ and $X'$ are the same and the result follows. Lemma 3: The circumcenter of triangle $DEF$ is $T$ This is a direct result of the fact that $T$ lies on the perpendicular bisectors of $DE$ and $EF$. Now to finish, $$2\angle DEF=\angle DTE=\angle ABC$$

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