Let's begin with the following useful lemma.

Lemma. Let points $F$ and $E$ be chosen on sides $AB$ and $AC$ of parallelogram $ABKC$ respectively, such that $BE = CF$. Then point $K$ is equidistant from lines $BE$ and $CF$ (see Fig. 1).

Рис. 1

Proof. Since $BK \parallel EC$ and $CK \parallel FB$, we have $S_{KBE} = S_{KBC} = S_{KFC}$. As $BE = CF$, it follows that the distances from point $K$ to lines $BE$ and $CF$ are equal. $\square$

Now let's proceed to the solution. Suppose the lines given in the condition form triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ (points are labeled as in Fig. 2).

Choose point $K$ such that $ABKC$ is a parallelogram; according to the lemma, point $K$ is equidistant from lines $BE_1, CF_1, BE_2$ and $CF_2$; therefore, there exists a circle centered at $K$ that is tangent to these lines at points $P_1, Q_1, P_2$ and $Q_2$ respectively. Then from the equality of tangent segments we obtain:

Рис. 2

$$\begin{aligned} BX_1 - CX_1 &= BP_1 + X_1P_1 - X_1Q_1 + CQ_1 = BP_2 + CQ_2 = \\ &= BP_2 - X_2P_2 + X_2Q_2 + CQ_2 = CX_2 - BX_2. \end{aligned}$$

Similarly, we get $CY_1 - AY_1 = AY_2 - CY_2$ and $AZ_1 - BZ_1 = BZ_2 - AZ_2$. Adding these three equalities yields the required equality of perimeters.