General Trema Construction with One Rotation
Suppose fixed points are located at the points ${F_1}$, ${F_2}$, and ${F_3}$, where $F_1$ is the fixed point for the rotation by 180° and $F_2F_3$ form a horizontal segment (the fixed points can always be rotated to achieve this).
The functions corresponding to the chaos game moves for these fixed points are $$ \begin{aligned} f_1(P) &= P + \tfrac32(F_1-P) = -\tfrac12P+\tfrac32F_1 \\ \\ f_2(P) &= P + \tfrac12(F_2-P) = \tfrac12P + \tfrac12F_2 \\ \\ f_3(P) &= P + \tfrac12(F_3-P) = \tfrac12P + \tfrac12F_3 \end{aligned} $$
Take $A = F_2$ and $B = F_3$. Let $C = f_1(F_2)$, i.e. $F_2$ is moved halfway towards $F_1$, then rotated 180° around $F_1$, and let $D = f_1(F_3)$, i.e. $F_3$ is moved halfway towards $F_1$, then rotated 180° around $F_1$. Then $$ \begin{aligned} C &= f_1(F_2) = F_2 + \tfrac32(F_1-F_2) = -\tfrac12F_2 + \tfrac32F_1 \\ D &= f_1(F_3) = F_3 + \tfrac32(F_1-F_3) = -\tfrac12F_3 + \tfrac32F_1 \\ \end{aligned} $$ and $$ \begin{aligned} \overrightarrow{DC} &= C - D = \tfrac12F_3 - \tfrac12F_2 \\ \overrightarrow{AB} &= B - A = F_3 - F_2 \end{aligned} $$
Therefore AB and DC are parallel, so $ABCD$ is a trapezoid with $|DC|= \tfrac12|AB|$.
The following GeoGebra app shows three such fixed points and the trapezoid formed by the calculations above. You can move the point $P$ around the four edges of the trapezoid, either manually by moving the point on the slider, or automatically by clicking on the Play button. As $P$ moves around the trapezoid, the green points $f_1(P)$, red points $f_2(P)$, and blue points $f_3(P)$ will trace where these points would occur while playing the Sierpinski gasket chaos game with the point $P$.
Notice that as $P$ moves counterclockwise around the edges of the trapezoid, the point $f_1(P)$ moves clockwise while points $f_2(P)$ and $f_3(P)$ move counterclockwise. The points trace out three copies of the trapezoid each scaled by 1/2, with the green one rotated by 180°. Their boundaries form the lines used in the trema construction.
The following image summarizes the coordinates of the vertices of the trapezoid in terms of the fixed points, as well as the trema construction for the trapezoid. Point $E$ is the midpoint of $AB$ and points $F$ and $I$ are the midpoints of the legs of the trapezoid so that segment $FI$ is at half the height of the trapezoid.
Because the opposite sides $AE$ and $DC$ are parallel and of equal length, $AECD$ is a parallelogram. For a similar reason, $EBCD$ is also a parallelogram. Therefore $EC$ and $AD$ are parallel, and $ED$ and $BC$ are also parallel. These observations are what were used in the proofs for the isosceles trapezoid, and can be used here in the same way to show that the trema calculation for the trapezoid divides the trapezoid into three self-similar pieces scaled by 1/2 with one piece rotated by 180°.
Use the following GeoGebra app to experiment with changing the locations of the three fixed points. As the points are moved, the three smaller trapezoids will continue to maintain the self-similarity scaling of 1/2 determined by the three red dashed lines in the trema construction.
