Solution:

All angles occurring in what follows are to be understood as oriented and modulo $180^{\circ}$; this makes it unnecessary to consider the relative position of the points involved. The feet of the perpendiculars from $B$, $X$ onto $A C$, $E F$ shall be denoted by $P$, $Q$.

Strategy. Show the congruence of the triangles $A B P$, $E X Q$. $(*)$

From this $B P = X Q$ will immediately follow, and hence the claim. The proof of $(*)$ itself is carried out by means of a well-known congruence criterion in three steps.

I. Since $\text{B P A E Q X 90}$, both triangles are right-angled.

II. Since $A E \| B D$, the cyclic quadrilateral $A B D E$ is an isosceles trapezoid and hence $A B = D E$. Furthermore $D E = X E$ by the construction of $X$, and therefore $A B = X E$, i.e. the hypotenuses agree.

III. As before we conclude from $B C \| D F$ that $B C D F$ is likewise an isosceles trapezoid. By repeated use of the inscribed angle theorem we now obtain $\text{D E C + C A B D A C + C A B D A B D F B C D F C E F C E X + X E F}$. Since $\text{D E C C E X}$, it follows that $\text{C A B X E F}$, or - stated differently - $\text{P A B X E Q}$.