Maximum area of a rectangle inscribed in a triangle is $1/2$ the area of triangle
Solution 1:
Hint:
The hyperbola whose asymptotes are $DA$ and $DC$ and passes through $F$ is tangent to $AC$.
Solution 2:
Let $AB=c$, $CD =h$, $FH=x$, $FG=y$, then it's easy to find that $hy+cx = ch$. By AM-GM, $2\sqrt{(hy)(cx)} \leq hy + cx = ch$, so $xy <= ch/4 = A/2$