Okay, I will prove that $\frac{A_1A_2}{BC} = \frac{B_1B_2}{CA} = \frac{C_1C_2}{AB}$ implies $\frac{C_1B_2}{YZ} = \frac{A_1C_2}{ZX} = \frac{B_1A_2}{XY}$. Then, the converse follows by analogy. Let P be a point such that the quadrilateral $C_2A_1A_2P$ is a parallelogram. Then, $C_2P = A_1A_2$ and $PA_2 = C_2A_1$ with directed segments (segments on parallel lines are directed equally). Moreover, $C_2P \parallel A_1A_2$ and $PA_2 \parallel C_2A_1$; in other words, $C_2P \parallel BC$ and $PA_2 \parallel ZX$. Now, $\frac{C_2P}{BC}=\frac{A_1A_2}{BC}=\frac{C_1C_2}{AB}=\frac{C_2C_1}{BA}$; on the other hand, $C_2P \parallel BC$ and $C_2C_1 \parallel BA$. This entails that triangles $C_2PC_1$ and BCA are homothetic, so that $PC_1 \parallel CA$, and $\frac{PC_1}{CA}=\frac{C_1C_2}{AB}$, i. e. $\frac{PC_1}{CA}=\frac{B_1B_2}{CA}$, and $PC_1 = B_1B_2$. From $PC_1 \parallel CA$, we have also $PC_1 \parallel B_1B_2$. Hence, the quadrilateral $PB_1B_2C_1$ is a parallelogram, and we get $PB_1 \parallel C_1B_2$ and $PB_1 = C_1B_2$. We rewrite $PB_1 \parallel C_1B_2$ as $PB_1 \parallel ZY$. On the other hand, $PA_2 \parallel ZX$, as we have seen before. Furthermore, it is obvious that $B_1A_2 \parallel YX$. Thus, the triangles $PB_1A_2$ and ZYX are homothetic, and we get $\frac{PB_1}{ZY}=\frac{B_1A_2}{YX}=\frac{PA_2}{ZX}$, or, in other words, $\frac{C_1B_2}{ZY}=\frac{B_1A_2}{YX}=\frac{C_2A_1}{ZX}$. Rewriting this, we get $\frac{C_1B_2}{YZ}=\frac{A_1C_2}{ZX}=\frac{B_1A_2}{XY}$. This proof is not really new, it is just the generalization of a similar proof for a problem proposed by Spain for the IMO 1985 long/shortlist: Quote: Let $A_1A_2$, $B_1B_2$, $C_1C_2$ be three equal segments on the sides of an equilateral triangle. Prove that in the triangle formed by the lines $B_2C_1$, $C_2A_1$, $A_2B_1$, the segments $B_2C_1$, $C_2A_1$, $A_2B_1$ are proportional to the sides in which they are contained. Of course, this problem is not only a special case of sam-n's result, but it is also equivalent to it by affine transformations. Darij