Area of a right triangle with right angle on a bounding square diagonal I'm struggling with a problem and can't find a way to approach it.
The problem is as follows:
A right triangle EFD is constructed inside a square ABCD, such that the right angle ∠FED  is located on the square's diagonal AC, another angle ∠EFD is on BC, and the third angle is on ∠CDA.
It is also known that EG = 2AE, and that AB=BC=CD=DA=1.
How can we find the area of the inner triangle EFD?

I tried several approaches - finding similar triangles containing the proportional segments of the diagonal, trying to find expressions for the triangle's sides based on AE and/or EG through the pythagoras theorem, and similar directions. So far I couldn't find an approach that seems to get anywhere...
Any pointers or general assistance will be greatly appreciated!
 A: Three steps are necessary.

*

*First, I asssumed accidentally that $\triangle DEF$ is isosceles, but we can prove this. Since quadrilateral $CDEF$ has two opposite right angles, it is cyclic, so draw its circumcircle. Since $\angle FCE = \angle ECD = 45^\circ$, the arcs $FE$ and $ED$ are equal, so the chords $FE$ and $ED$ are also equal.


*We spot that in the diagram below, the two red triangles are congruent. (Their angles are equal, and their hypotenuses are also equal.) One way to phrase this: if we project $E, F, G$ down onto $AD$ to get $E', F', G'$, then we don't just have $E'G' = 2AE'$ (as we are given) but also $E'F' = AE'$ (because both are equal to the short side of a red triangle). If $x = AE'$, then $$AE' = x, \; E'F' = x,\; F'G' = x,\; G'D = 1-3x.$$


*In the blue triangle, $CG$ is an angle bisector. Therefore $FC : CD = FG : GD$ by the angle bisector theorem. We have $FC : CD = F'D : CD = (1-2x) : 1$, while $FG : GD = F'G' : G'D = x : (1-3x)$. So we can solve $\frac{x}{1-3x} = 1-2x$ for $x$.

Once we know $x$, we are basically done; the blue triangle has area $1-3x$, the two red triangles together have area $x(1-x)$, and the rectangle to their left has area $x$. So the leftover area for $\triangle DEF$ is $1 - (1-3x) - x(1-x) - x = x(x+1)$.
A: I ended up brute forcing a solution on Desmos and found that when point $E$ is at $x=\dfrac1{3+\sqrt3}$, we have $EG=2AE$, and the area of $DEF$ is $\dfrac13$.

We first create the square $ABCD$ and the diagonal $AC$, and draw a line passing through $D$ and $E$ with a parameter $x_1$ that maps the $x$-coordinate of $E$.  A perpendicular is dropped onto $DE$ such that it passes through $BC$, and point $F$ is found.  After that, point $G$ can be found by solving a system of two linear equations; $G$ is located at the point of intersection of $DF$ and $AC$, which both have an algebraic representation.
We can now solve an equation for the parameter $x_1$ such that $EG=2AE$, and since $DEF$ is a $1-1-\sqrt2$ triangle, finding the area becomes trivial.
