The question #To prove two angles are equal when some angles are supplementary in a parallelogram has been solved. In the process of solving it, I found it is not that easy to draw the corresponding diagram..
Let’s start from P. Through it, 4 distinct, non-collinear rays PQ, PR, PS and PT are drawn with $\angle QPR$ and $\angle SPT$ are supplementary. A and B are pre-sectected points on PQ and PR respectively such that AB is of fixed length.
We are then supposed to find C and D on PR and PT respectively such that ABCD is parallelogram. [I don’t think the translation of a line is an acceptable Euclidean construction.]
Two questions:-
1) Can we prove that there always exist (at least one or may be only one) such a parallelogram?
2) If yes, what are the construction steps?
1) Such a parallelogram does not always exist.
In this diagram, $\angle QPR$ and $\angle SPT$ are both right angles, so they are obviously supplementary. $m\angle RPS=30°$ and $\overline{AB}$ is drawn so that $\triangle ABP$ is isosceles. We can see from the orientations of $\overline{AB}$ and $\angle SPT$ that no points $C$ on $\overrightarrow{PS}$ and $D$ on $\overrightarrow{PT}$ can be chosen such that $\overline{AB}\parallel\overline{CD}$.
2) If the orientations of $\overline{AB}$ and $\angle SPT$ permit it then we can construct the points $C$ and $D$. In this diagram I have chosen different points $A$ and $B$ to allow this.
Choose an arbitrary point on $\overrightarrow{PS}$: I chose $S$ for simplicity. Construct point $U$ on $\overrightarrow{PT}$ so that $\overline{SU}\parallel\overline{AB}$. Draw the circle with center $U$ and radius $AB$. Mark the point $V$ at the intersection of the circle with line $\overleftrightarrow{SU}$.
Construct point $C$ on ray $\overrightarrow{PS}$ so that $\overline{CV}\parallel\overline{PU}$. Construct point $D$ on $\overrightarrow{PU}$ so that $\overline{CD}\parallel\overline{AB}$.
Then quadrilateral $ABCD$ is the desired parallelogram. That can be seen by examining parallelogram $CVUD$ and noting that $\overline{UV}=\overline{AB}$.