Answer. Yes, always.
If $\Delta ABC = \Delta A_1B_1C_1$, then the construction method is obvious. Otherwise, we may assume that $AB \neq A_1B_1$ (let's suppose $AB < A_1B_1$ for definiteness).

Construct triangle $A'B'C$ such that $AB \parallel A'B'$, $A'B' = A_1B_1$, $B'C = B_1C_1$, $CA' = C_1A_1$ (see Fig. 23; this construction is easily carried out, for example, using the equality $\angle BCB' = \angle ABC + \angle A_1B_1C_1$). Then $ABB'A'$ is a trapezoid. Let $MN$ be its midline, and $P$ the intersection point of the extensions of the lateral sides.

Construct a circle $\omega$ on segment $PC$ as diameter. Since $AB < A_1B_1 = A'B'$, $S_{ABC} = S_{A_1B_1C_1} = S_{A'B'C'}$, the distance from point $C$ to line $AB$ is greater than the distance from $C$ to line $A'B'$, so points $P$ and $C$ lie on opposite sides of line $MN$, and therefore $\omega$ intersects line $MN$. Let $K$ be one of the intersection points (in the figure it lies on segment $MN$, but our reasoning does not depend on this).

Draw line $PK$ to intersect lines $AB$ and $A'B'$ at points $X$ and $Y$, respectively. Then $\frac{AX}{XB} = \frac{A'Y}{YB'}$, so $S_{XBC} = S_{YB'C'}$. Moreover, $XK = KY$, and angle $PKC$ is right as it subtends the diameter in circle $\omega$. Thus, $CK$ is the perpendicular bisector of segment $XY$, and $CX = CY$.

On the extension of segment $XC$ beyond point $C$, take point $Z$ such that $XC = CZ$. Construct triangles $A_2CZ$ and $B_2CZ$ congruent to triangles $A'CY$ and $B'CY$, respectively (see Fig. 24). Then $\triangle A_2B_2C = \triangle A'B'C' = \triangle A_1B_1C_1$. We show that $AA_2 \parallel BB_2$ (then we can shift triangle $A_2B_2C$ along line $AA_2$, obtaining the required one). Since $CX = CZ$, $S_{ABC} = S_{A_2B_2C}$ and $S_{XBC} = S_{ZB_2C}$, it follows that $S_{XAC} = S_{ZA_2C}$, the distances from points $A$ and $A_2$ to line $XZ$ are equal, and the distances from points $B$ and $B_2$ to line $XZ$ are also equal. Therefore, $AA_2 \parallel XZ \parallel BB_2$, as required.