First we prove the equality
$$\frac{B_1 E}{C_1 D} = \frac{B_1 A_1}{C_1 A_1}.$$
Since the quadrilateral $XB_1CE$ is cyclic, $\angle XB_1E = \angle XCE$. Therefore, the triangle $BB_1E$ is similar to the triangle $BCX$. Hence,
$$\frac{B_1 E}{X C} = \frac{B B_1}{B C}. \qquad (1)$$
Similarly one can prove the equality
$$\frac{C_1 D}{X B} = \frac{C C_1}{B C}. \qquad (2)$$
Hence, dividing equation (1) by equation (2), we get
$$\frac{B_1 E}{C_1 D} = \frac{X C}{X B} \cdot \frac{B B_1}{C C_1}.$$
Applying Menelaus' theorem to the triangle $XB_1C$ and the line $AB$ we obtain
$$\frac{X C_1}{C C_1} \cdot \frac{A C}{A B_1} \cdot \frac{B B_1}{X B} = 1. \qquad (3)$$
Further, the same arguments applied to the triangle $CC_1B_1$ and the line $AX$ imply the equality
$$\frac{C_1 A_1}{B_1 A_1} \cdot \frac{A B_1}{A C} \cdot \frac{X C}{X C_1} = 1. \qquad (4)$$
Thus, multiplying equalities (3) and (4), we obtain
$$\frac{B_1 A_1}{C_1 A_1} = \frac{X C}{X B} \cdot \frac{B B_1}{C C_1} = \frac{B_1 E}{C_1 D}.$$

$$\frac{B_1F}{FD} = \frac{B_1E}{C_1D} = \frac{B_1A_1}{C_1A_1},$$
so $A_1F \parallel C_1D$.
Suppose that the lines $B_1E$ and $C_1D$ intersect at point $Y$. To prove the statement of the problem it is sufficient to show that the lines $YA_1$, $B_1D$ and $C_1E$ intersect at the same point. Let us prove the equality $YD = YE$. Indeed, since the quadrilaterals $XB_1CE$ and $XC_1BD$ are cyclic,
$$\angle DEY = \angle CEB_1 = \angle CXB_1 = \angle C_1XB = \angle C_1DB = \angle EDY.$$
Therefore, the triangle $YDE$ is isosceles. Since
$$\frac{YE}{B_1E} \cdot \frac{B_1A_1}{C_1A_1} \cdot \frac{C_1D}{YD} = \frac{B_1A_1}{C_1A_1} \cdot \frac{C_1D}{B_1E} = 1,$$
so Ceva's theorem implies that the lines $YA_1$, $B_1D$ and $C_1E$ intersect at the same point.