First solution. Let the line $EF$ meet the circle $C_1BE$ again at $T$. Then
$$\angle (TC_1, C_1B) = \angle (TE, EB) = \angle (FE, EA) = \angle (FA_1, A_1A),$$
so $TC_1$ and $FA_1$ are parallel. Similarly,
$$\angle (TB, BC_1) = \angle (TE, EC_1) = \angle (FE, EC_1) = \angle (FC, CC_1) = \angle (FD, DA_1),$$
so $TB$ and $FD$ are parallel.
Thus, the corresponding sides of the triangles $A_1DF$ and $C_1BT$ are parallel. Since the vectors $\overrightarrow{C_1B}$ and $\overrightarrow{A_1D}$ are counter-directed, these triangles are homothetical under a homothety of negative ratio.
The centre of this homothety $O$ lies on the lines $BD$, $A_1C_1$, and $TF = EF$, so the three are concurrent.

Second solution. Let the lines $AB$ and $CD$ meet at $S$, and let the lines $A_1F$ and $C_1E$ meet at $K$. Since
$$\begin{align*} \angle(EK, FK) &= \angle(EC_1, EF) + \angle(EF, A_1F) = \angle(CC_1, CF) + \angle(EA, AA_1) \\ &= \angle(BC, FS) + \angle(ES, BC) = \angle(ES, FS), \end{align*}$$
the points $E$, $F$, $S$, and $K$ are concyclic. Thus $\angle(KS, KF) = \angle(ES, EF) = \angle(EA, EF) = \angle(A_1A, A_1F)$, showing that $KS$ is parallel to both $AD$ and $BC$.
This shows the triangles $A_1DF$ and $C_1BE$ in perspective from the ideal point of the three parallel lines, so the lines $A_1C_1$, $BD$ and $EF$ are concurrent, by Desargues' theorem; since the segments $A_1C_1$ and $BD$ have a non-empty intersection, the point of concurrency is not ideal.