On any given line segment, a parallelogram can be constructed in a given angle the same size as any given triangle.
Let $AB$ be the given line segment, $C$ be the given triangle and let $D$ be the given angle.
From Construction of Parallelogram Equal to Triangle in Given Angle, construct the parallelogram $BEFG$ equal to $C$ such that $\angle EBG = \angle D$.
Place it so that $BE$ is in a straight line with $AB$.
Let $FG$ be produced to $H$, and draw $AH$ parallel to $BG$.
Then join $HB$.
We have that $HF$ falls on the parallel lines $AH$ and $EF$.
So from Parallel Implies Supplementary Interior Angles, $AHF$ and $HFE$ are supplementary.
So $\angle BHG + \angle GFE$ is less than two right angles.
From the Fifth Postulate, $HB$ and $FE$ will meet if produced. So let them meet at $K$.
Draw $KL$ parallel to $EA$, and produce $HA$ and $GB$ to the points $L$ and $M$.
Then $HLKF$ is a parallelogram, and from Complements of Parallelograms are Equal, $ABML$ has the same area as $BEFG$.
But $BEFG$ has the same area as $\triangle C$, and so by common notion 1, so does $ABML$.
From Two Straight Lines make Equal Opposite Angles, $\angle GBE = \angle ABM$, while $\angle GBE = \angle D$.
Therefore $ABML$ is the required parallelogram.
This is Proposition 44 of Book I of Euclid's The Elements.
There is a suggestion that this result is the work of Pythagoras.