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.
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$$\frac{1}{2}(2)(8)\sin 135^o = \frac{1}{2}(2)(8)\frac{\sqrt 2}{2}=4\sqrt 2$$
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 $\triangle BEC=4\sqrt{2}+16-16=4\sqrt{2}$
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 $CQ\times BE/2=4\sqrt2$
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: $$ \text{area of $BEC$}={1\over2}ay={a^2\over4}-17=4\sqrt2. $$
(I am applying a method I used in a later problem; the diagram is scavenged [so I wouldn't have to re-draw it completely], so the proportions are not-to-scale.)
Superimpose a horizontal reflection of square $ \ ABCD \ $ onto the original square. If we call $ \ s \ $ the length of a side of the square, we produce two isosceles trapezoids $ \ ABEE' \ $ and $ \ DCEE' \ \ , \ $ both with parallel sides of lengths $ \ s \ $ and $ \ u \ \ , \ $ the former with altitude $ \ h \ $ and the later with altitude $ \ s - h \ \ . \ $ Trapezoid $ \ DCEE' \ $ has a "slant height" of $ \ 8 \ \ , \ $ while trapezoid $ \ ABEE' \ $ has slant height of $ \ 2 \ $ and diagonals of length $ \ 6·\sqrt2 \ \ . \ $ The line containing $ \ EE' \ $ meets the sides of the square at $ \ G \ $ and $ \ I \ \ , \ $ thus $ \ EG \ $ and $ \ E'I \ $ have length $ \frac{s \ - \ u}{2} \ $ and are the altitudes of congruent triangles $ \ \triangle BEC \ $ and $ \ \triangle AE'D \ \ . \ $ Marking the line of trapezoid altitudes $ \ EF \ \ $ with length $ \ h \ $ and $ \ EH \ $ with length $ \ s - h \ \ , \ $ we have the length of segment $ \ AF \ $ as $ \frac{s \ + \ u}{2} \ \ . \ $ We will call $ \ \theta \ $ the measure of $ \ \angle BEC \ \ . $
The "Pythagorean" Theorem allows us to write $$ (s - h)^2 \ + \ \left( \frac{s \ - \ u}{2} \right)^2 \ \ = \ \ 8^2 \ \ \ , $$ $$ h^2 \ + \ \left( \frac{s \ - \ u}{2} \right)^2 \ \ = \ \ 2^2 \ \ \ , \ \ \ h^2 \ + \ \left( \frac{s \ + \ u}{2} \right)^2 \ \ = \ \ (6·\sqrt2)^2 \ \ . $$ If we subtract the second of these equations from the third, we obtain $$ \left( \frac{s \ + \ u}{2} \right)^2 \ - \ \left( \frac{s \ - \ u}{2} \right)^2 \ \ = \ \ su \ \ = \ \ 72 \ - \ 4 \ \ = \ \ 68 \ \ . $$
The area of triangles $ \ \triangle BEC \ $ and $ \ \triangle AE'D \ $ are $ \ \mathcal{A}_{\triangle} \ = \ \frac12·s· \frac{s - u}{2} \ = \ \frac14·(s^2 - su) \ \ $ and also, using included angle $ \ \angle BEC \ \ , \ \ \mathcal{A}_{\triangle} \ = \ \frac12·2· 8·\sin \theta \ = \ 8·\sin \theta \ \ . $
The area of $ \ \square ABCD \ $ is the sum of $$ \mathcal{A}_{\square} \ \ = \ \ s^2 \ \ = \ \ \mathcal{A}_{ABEE'} \ + \ \mathcal{A}_{DCEE'} \ + \ 2·\mathcal{A}_{\triangle} $$ $$ = \ \ \left( \frac{s \ + \ u}{2} \right)·h \ + \ \left( \frac{s \ + \ u}{2} \right)·(s - h) \ + \ 2·8·\sin \theta \ \ = \ \ \frac12·s^2 \ + \ \frac12·su \ + \ 16·\sin \theta $$ $$ \Rightarrow \ \ \frac12·s^2 \ \ = \ \ 34 \ + \ 16·\sin \theta \ \ . $$
But in applying the Law of Cosines to $ \ \triangle BEC \ \ , \ $ we also find $ \ s^2 \ = \ 8^2 + 2^2 - 2·8·2· \cos \theta \ = \ 68 - 32·\cos \theta \ \ . \ $ Hence, $$ s^2 \ \ = \ \ 68 \ - \ 32·\cos \theta \ \ = \ \ 68 \ + \ 32·\sin \theta \ \ \Rightarrow \ \ -\cos \theta \ \ = \ \ \sin \theta $$ $$ \Rightarrow \ \ \tan \theta \ \ = \ \ -1 \ \ \Rightarrow \ \ \theta \ \ = \ \ 135º \ \ , $$ since $ \ \angle BEC \ $ is one angle of a triangle. We conclude that $ \ \boxed{ \ \mathcal{A}_{\triangle} \ = \ 8·\sin 135º \ = \ 8·\frac{\sqrt2}{2} \ = \ 4·\sqrt2 \ } \ \ . $
[As with the linked problem, it is well that we do not work directly with $ \ s \ \ , \ $ which we now see is given by $ \ s^2 \ = \ 68 + 16· \sqrt2 \ \Rightarrow \ s \ = \ 2·\sqrt{ \ 17 + 4 \sqrt2 } \ \ , \ $ a number without a simpler radical expression. We can check, however, that $$ \mathcal{A}_{\triangle} \ \ = \ \ \frac14·(s^2 - su) \ \ = \ \ \frac14·( \ [68 + 16· \sqrt2] \ - \ 68 \ ) \ \ = \ \ 4· \sqrt2 \ \ . \ ] $$
