Answer: The line passing through $A$ parallel to the line $KF$ where $K$ is the point on $AC$ such that $\angle AA_1K = 90^\circ$ (excluding $A$, and the intersection points of this line with the line $BC$ and the lines passing through $F$ and $D$ parallel to $AC$ and $AB$, respectively).
Choose the points $L$ and $K$ on the lines $AB$ and $AC$ such that the line $KL$ passes through $A_1$ perpendicular to $AA_1$. The quadrilateral $LDKF$ is a parallelogram since $A_1$ is the midpoint of the segments $DF$ and $LK$. Denote the point of intersection of the lines $B_1D$ and $C_1F$ by $S$. Since the points $AL \cap SD = B_1$, $LK \cap DF = A_1$ and $KA \cap FS = C_1$ are concurrent, from the Desargues's theorem it follows that the triangles $ALK$ and $SDF$ are centrally perspective with the center — point of intersection of $AS$, $LD$ and $KF$. The lines $LD$ and $KF$ are parallel, so the point $S$ belongs to the line $s$ passing through $A$ parallel to $KF$ and $LD$.
On the other hand, take an arbitrary point $S$ (different from the four points mentioned in the answer) on $s$ and let $B_1 = SF \cap AC$ and $C_1 = SD \cap AB$. Since the triangles $ALK$ and $SDF$ are centrally perspective, from the Desargues's theorem it follows that they are axially perspective with the axis — line passing through $C_1$, $A_1$ and $B_1$. This axis is the line $l$ from the problem condition.