Area of a right triangle with right angle on a bounding square diagonal

Question

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!

Answer 1

Three steps are necessary.

1. 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.

2. 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 $$AE' = x, \; E'F' = x,\; F'G' = x,\; G'D = 1-3x.$$

3. 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 $1 - (1-3x) - x(1-x) - x = x(x+1)$.

Answer 2

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.