Selection and Training Session

First we prove the equality $$\frac{B_{1}E}{C_{1}D}=\frac{B_1{A}_1}{C_1{A}_1}.$$ Since the quadrilateral $X{B}_{1}CE$ is cyclic, $\angle X{B}_{1}E=\angle XCE$. Therefore, the triangle $B{B}_{1}E$ is similar to the triangle $BCX$. Hence, $$\frac{B_{1}E}{XC}=\frac{B{B}_1}{BC}.(1)$$ Similarly one can prove the equality $$\frac{C_{1}D}{XB}=\frac{C{C}_1}{BC}.(2)$$ Hence, dividing equation (1) by equation (2), we get $$\frac{B_{1}E}{C_{1}D}=\frac{XC}{XB}\cdot \frac{B{B}_1}{C{C}_1}.$$ Applying Menelaus' theorem to the triangle $X{B}_{1}C$ and the line $AB$ we obtain $$\frac{X{C}_1}{C{C}_1}\cdot \frac{AC}{A{B}_1}\cdot \frac{B{B}_1}{XB}=1.(3)$$ Further, the same arguments applied to the triangle $C{C}_1{B}_1$ and the line $AX$ imply the equality $$\frac{C_1{A}_1}{B_1{A}_1}\cdot \frac{A{B}_1}{AC}\cdot \frac{XC}{X{C}_1}=1.(4)$$ Thus, multiplying equalities (3) and (4), we obtain $$\frac{B_1{A}_1}{C_1{A}_1}=\frac{XC}{XB}\cdot \frac{B{B}_1}{C{C}_1}=\frac{B_{1}E}{C_{1}D}.$$



$$\frac{B_{1}F}{FD}=\frac{B_{1}E}{C_{1}D}=\frac{B_1{A}_1}{C_1{A}_1},$$ so $A_{1}F\parallel C_{1}D$. Suppose that the lines $B_{1}E$ and $C_{1}D$ intersect at point $Y$. To prove the statement of the problem it is sufficient to show that the lines $Y{A}_1$, $B_{1}D$ and $C_{1}E$ intersect at the same point. Let us prove the equality $YD=YE$. Indeed, since the quadrilaterals $X{B}_{1}CE$ and $X{C}_{1}BD$ are cyclic, $$\angle DEY=\angle CE{B}_1=\angle CX{B}_1=\angle C_{1}XB=\angle C_{1}DB=\angle EDY.$$ Therefore, the triangle $YDE$ is isosceles. Since $$\frac{YE}{B_{1}E}\cdot \frac{B_1{A}_1}{C_1{A}_1}\cdot \frac{C_{1}D}{YD}=\frac{B_1{A}_1}{C_1{A}_1}\cdot \frac{C_{1}D}{B_{1}E}=1,$$ so Ceva's theorem implies that the lines $Y{A}_1$, $B_{1}D$ and $C_{1}E$ intersect at the same point.