it seems we can use analytic, any better way?

We'll call the vertex A, B and C such that $ |AB|= {3 \over 2}, |AC| = {{\sqrt 5}\over 2}$ so $ |BC| = \sqrt 2$. Let $ S$ be the area of the overlapping portions of the folded triangle, and $ S_0 = {1 \over 4}$ the "starting" value of S, when the folding line contains vertex C (or the hight $ h = 1$, relative to the base AB). If we now displace the folding line toward the vertex B by an amount of $ x$, we'll add to $ S$ the area of the trapeze with bases $ h, (h - tan \angle ABC)$ and height $ x$, by one side, but in the other side, subtract the area of the rectangular (thanks to $ \angle ABC = {\pi \over 4}$) triangle with cathetus $ x. sec \angle ABC$ adjacent to $ \angle BCA$. So $ S = {1 \over 4} + x - {{x^2} \over 2} - 3x^2$, whose maximum of $ {9 \over {28}}$ occurs when $ x = {1 \over 7}$.