Euclid claims in I.45 of his Elements to show how to "construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle." In modern terms, he is saying that he will show how to construct a parallelogram that (a) has the same area as a given polygon and (b) has an interior angle whose measure is equal to that of a given angle. (A link to Euclid's claim and construction may be found here: https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI45.html)
However, Euclid's construction is actually only for the case where the polygon is a quadrilateral. I understand how to do this, but I would like to know how to do the construction for an arbitrary polygon, or, at least, an arbitrary pentagon.
Here is my understanding of the case where the polygon is a quadrilateral:
In case it matters/helps, I am trying to do the construction in GeoGebra.
If you translate the vertex of a triangle parallel to the opposite side you get a new triangle with the same area.
Given a polygon, look at the triangle formed by three consecutive vertices. You can translate the middle vertex parallel to the diagonal formed by the other two vertices until one of the sides disappear. This is, draw the parallel to the diagonal passing through the middle vertex. Then move the vertex to the intersection of this parallel and the next side (formed by one of the vertices following the three given vertices). This turns your polygon into one with the same area but one less side.
Through this process you can get to a triangle with the same area as the polygon. With a triangle draw a parallel to one side passing through the mid point of the two other sides and extend it until it meets a parallel to one of those sides passing through the opposite vertex. This produces a rectangle with the same area as the triangle.