$ABCD$ is a square, point $E$ lies inside the square such that $BE=2$, $AE=6\sqrt{2}$, $CE=8$, calculate the area of $\triangle BEC$. The question is as stated in the title, in the figure given below, find the area of $\triangle BEC$. I must admit this was a challenging problem and the solution that I came up with (which will also be posted as an answer) is pretty complicated and "messy". So, I'd like to see if there are any better approaches that may be simpler as well.

 A: This is my own approach. I'll add an explanation below as well!

Here's how I go about it:
1.) Rotate $\triangle AEB$ by $90°$ counterclockwise to form a new triangle $\triangle CFB$ that is congruent to $\triangle AEB$. Notice that $\angle EBF=90$ and $EB=BF=2$. Connect $E$ and $F$ via segment $EF$. Notice that single $\triangle EBF$ is am isosceles right triangle, segment $EF=2\sqrt{2}$.
2.) Onto $\triangle ECF$, notice that all of its sides $(8, 2\sqrt{2}, 6\sqrt{2})$ are Pythagorean triples, with $6\sqrt{2}$ being the longest side, this proves that $\angle FEC=90$, since we already established that $\triangle EBF$ is an isosceles right triangle, we can say that $\angle BEF=45$. Therefore, $\angle BEC=90+45=135$.
3.) Extend segment $BE$ to meet at point $G$ and connect $G$ with $C$ via $GC$ such that $\angle CGE=90$. Since $\angle GCE=\angle GEC=45$, we can conclude that $\triangle GEC$ is also an isosceles right triangle with $EC=8$. This means that $GE=GC=4\sqrt{2}$ ($EC$ divided by $\sqrt{2}$).
4.) The area of $\triangle BEC$=$A(\triangle GBC)-A(\triangle GEC)$. Therefore, area of $\triangle BEC=4\sqrt{2}+16-16=4\sqrt{2}$
A: 
Extend $BE$, $AP\perp BE$ and $CQ\perp BE$. $EP=x$ and $PQ=y$. So we get
\begin{cases}
(2+x+y)^2+x^2=(6\sqrt 2)^2\qquad (1)\\
(x+y)^2+(2+x)^2=8^2\qquad (2)
\end{cases}
From $(1)-(2)$ we get $y=2$. Plug it in to $(2)$ we get
\begin{equation*}
(x+2)^2+(2+x)^2=8^2
\end{equation*}
So $x+2=4\sqrt2$. The area of $\triangle BEC$ is $CQ\times BE/2=4\sqrt2$
A: For a straightforward solution, let $a$ be the length of a side of the square, $x$ the distance of $E$ from $AB$ and $y$ the distance of $E$ from $BC$. The given distances lead then to three equations:
$$
\cases{
x^2+y^2=4 \\
(a-x)^2+y^2=64\\
(a-y)^2+x^2=72
}
$$
Subtracting the first equation from the last two, one gets:
$$
x={a\over2}-{30\over a},\quad
y={a\over2}-{34\over a}
$$
and plugging these into the first equation one can solve for $a^2$:
$$
a^2=68+16\sqrt{2}.
$$
Finally:
$$
\text{area of $BEC$}={1\over2}ay={a^2\over4}-17=4\sqrt2.
$$
A: 
The figure is obtained by putting $\Delta BCE$ and $\Delta BAE$ together, with $BA$ overlapping $BC$.
Note that $$\angle CBE + \angle BEA = 90^o \implies \angle E_1BE_2=90^o$$
By Pythagoras theorem, $E_1E_2=\sqrt 8$
Noting that $$E_1A_C^2+E_1E_2^2=8+64=72=E_2A_C^2$$
By Converse of Pythagoras' Theorem, $\angle E_2E_1A_C=90^o$
Thus $\angle BE_1A_C =135^o$
Hence the required area is$$\frac{1}{2}(2)(8)\sin 135^o = \frac{1}{2}(2)(8)\frac{\sqrt 2}{2}=4\sqrt 2$$
