A recent question asked about finding the ratio of the bases for the following isosceles trapezoid:
That problem has been solved, obtaining a result of $|CD|/|AB|=1-1/\sqrt{2}$. What I'm curious how one would construct this figure in the first place. The solution to the original question offers some hints. Let $E$ be the central point, and let $h_1,h_2$ be the lengths of the altitudes drawn from $E$ to $AB,CD$ respectively. Then the areas of the triangles $\triangle ABE,\triangle CDE$ require
\begin{align} \frac12 h_1|AB|=4&\implies h_1=\frac{8}{|AB|}\;,\\ \frac12 h_2 |CD|=2&\implies h_2 =\frac{4}{|CD|}=\frac{4}{1-1/\sqrt{2}}\frac{1}{|AB|}=\frac{4}{|AB|}(2+\sqrt{2}) \end{align} This determines the vertical position, but still leaves the horizontal position of $E$. Is there a straightforward way to complete this construction? (Graphical demonstrations encouraged.)
Since $ABCD$ is isosceles, $\angle GEI=\angle GFH$. The two triangles are similar with ratio $IG:GH=5:3$.
So after constructing $EF$ parallel to the pair of parallel sides, divide it into ratio $5:3$.