Construction of Parallelogram in Given Angle Equal to Given Polygon
Join $DB$, and construct the parallelogram $FGHK$ equal in size to $\triangle ABD$, in $\angle HKF = \angle E$.
We now need to show that $KFLM$ is the required parallelogram.
By common notion 1, $\angle HKF = \angle GHM$ as both are equal to $\angle E$.
Add $\angle KHG$ to each, so as to make $\angle FKH + \angle KHG = \angle KHG + \angle GHM$.
So from Two Angles making Two Right Angles make Straight Line, $KH$ is in a straight line with $HM$.
From Parallel Implies Equal Alternate Interior Angles, $\angle MHG = \angle HGF$.
Add $\angle HGL$ to each, so as to make $\angle MHG + \angle HGL = \angle HGF + \angle HGL$.
So from Two Angles making Two Right Angles make Straight Line, $FG$ is in a straight line with $GL$.
From Parallelism is Transitive, as $KF \| HG$ and $HG \| ML$, it follows that $KF \| ML$.
Similarly, from common notion 1, $KF = ML$.
Therefore $KFLM$ is a parallelogram.
But the area of $KFLM$ equals the combined areas of $FGHK$ and $GLMH$, which are equal to the combined areas of $\triangle ABD$ and $\triangle BCD$.
This is Proposition 45 of Book I of Euclid's The Elements.
Note that this technique can be expanded for a polygon with any number of sides, merely by dividing the polygon up into as many triangles as it takes.