What is the measure of the $\measuredangle EAD$ where $E$ is outside the square $ABCD$? For reference:

Let the squares $ABCD$ and $FGDE$ such that E, G and C are collinear, $GE = GC$, Calculate the measure of the $\measuredangle EAD$ where $E$ is outside the square $ABCD$.(answer: 18.5 $^\circ$)

My progress..
I was able to draw the figure but I don't know there is some restriction...in geogebra the solution matches

 A: Draw $AJ \parallel CE$ such that $AJ = CE$. Use symmetry about diagonal $AC$.

$\triangle ACE$ is a well known right triangle with perpendicular sides in the ratio $1:2$. The angles are $26.5^\circ$ and $63.5^\circ$.
So, $\angle EAD = \angle 63.5^\circ - 45^\circ = 18.5^\circ$
Here is another observation. Points  $A, B, J, C, D$ and $E$ are all concyclic with center at $O$.
Here is a complete diagram with $9$ squares.

A: Clearly, triangles $DGC$ and $DEA$ are equal, since they are related by a $90$ degree rotation. Therefore, $\theta=DCG$.
Call $L=DC$ the side of the large square, and $l$ the side of the small square. By the law of cosines in triangle $DEC$ ($DEC=45$ degrees and $EC=2EG=2\sqrt{2}l$)
$$
L^2=l^2+(2\sqrt{2}l)^2-2l\cdot 2\sqrt{2}l\cdot\frac{\sqrt{2}}{2}
$$
which implies $L^2=5l^2$ hence $L=\sqrt{5}l$. By the law of sines in the same triangle
$$
\frac{l}{\sin\theta}=\frac{\sqrt{5}l}{\sqrt{2}/2}
$$
which gives
$$
\sin\theta=\frac{\sqrt{10}}{10}
$$
Therefore, your angle is $\theta=\arcsin \frac{\sqrt{10}}{10}\approx 18.5$ degrees.
A: Since $\measuredangle DEC=\measuredangle DAC=45^{\circ},$ we see that $DEAC$ is cyclic, which says $\measuredangle EAD=\measuredangle ECD.$
Now, let $DC=1$ and $DG=a$.
Thus, since $DG$ is a median of $\Delta EDC$, we obtain:
$$a=\frac{1}{2}\sqrt{2a^2+2\cdot1^2-(2\sqrt2a)^2},$$ which gives $a=\frac{1}{\sqrt5}.$
Id est, by law of cosines for $\Delta DGC$ we obtain:
$$\theta=\arccos\frac{GC^2+DC^2-DG^2}{2DC\cdot GC}=\arccos\frac{\frac{2}{5}+1-\frac{1}{5}}{2\sqrt{\frac{2}{5}}}=\arccos\frac{3}{\sqrt{10}}.$$
A: Here is another interpretation. Perform a $90^{\circ}$ clock-wise rotation around  $D$. Then, since $ABCD$ and $EFGD$ are squares with the center of rotation $D$ as a common vertex, $A$ is rotated to $C$ and $E$ is rotated to $G$. Consequently, triangle $ADE$ is rotated to triangle $CDG$ which means that $$\angle \, EAD = \angle \, GCD = \theta$$ Let $K$ be the intersection point of the diagonals $EG$ and $DF$ of the square $EFGD$. Then
$$KD = KG = KE = \frac{1}{2} GE = \frac{1}{2} GC$$ Since $C, G, E$ are collinear, then triangle $CDK$ is right-angled ($\angle \, CDK = 90^{\circ}$) with ratio
$$\frac{KD}{KC}  = \frac{KD}{KG + GC} = \frac{KG}{KG + 2KG} = \frac{1}{3}$$
Consequently, in the right angle triangle $CDK$
$$\tan(\theta) = \tan\big(\angle\, GCD\big) = \frac{KD}{KC} = \frac{1}{3}$$
Therefore, $$\angle \, EAD = \angle\, GCD = \theta = \arctan\left(\frac{1}{3}\right) = 18.435^{\circ}$$
