All angles appearing below are to be understood as oriented and modulo $180^{\circ}$. This makes considerations of the relative positions of the involved points unnecessary. The feet of the perpendiculars from $B, X$ to $A C, E F$ are denoted by $P, Q$.

Strategy. Show the congruence of triangles $ABP$ and $EXQ$. (*)
From this, $BP = XQ$ will immediately follow, and thus the claim will be proven. The proof of (*) itself proceeds in three steps using a known congruence theorem.

I. Since $\text{BPA EQX 90}$, both triangles are right-angled.
II. Because $AE \| BD$, the cyclic quadrilateral $ABDE$ is an isosceles trapezoid, and thus $AB = DE$. Furthermore, $DE = XE$ by the construction of $X$, and therefore $AB = XE$, i.e., the hypotenuses are equal.
III. As before, we conclude from $BC \| DF$ that $BCDF$ is also an isosceles trapezoid. By repeated application of the inscribed angle theorem, we now get $\text{}$ DEC + $\text{}$ CAB $\text{}$ DAC + $\text{}$ CAB $\text{}$ DAB $\text{}$ DFB $\text{}$ CDF $\text{}$ CEF $\text{}$ CEX + $\text{}$ XEF. From $\text{DEC CEX}$, it follows that $\text{CAB XEF}$, or, in other words, $\text{- PAB XEQ}$.

Remark. Discussions of positional relationships have not been taken into account either positively or negatively in the evaluation.