Square Geometry Problem 
http://imgur.com/a/5BYiH
I have a question related to geometry.  The problem is shown in the image above. If you have an idea related to the solution, I will be happy.
 A: Extend your picture to the following picture:

where $EFGH$ is a square. Then triangle $EFC$ is a right triangle with $$EF = ED + DF = ED + AE = 6 + 2 = 8$$ and $FC = 6$. By Pythagoras' theorem
$$EC^2 = EF^2 + FC^2 = 8^2 + 6^2 = 64 + 36 = 100$$ so $EC = 10$.
A: There are several ways to attack this problem, but the simplest involves literally thinking outside the box.
Construct the line $l$ through $C$ parallel to $AE$ and extend $EB$ to meet $l$ at $F$.  $\angle FBC$ is complementary to $\angle ABE$ and $\angle CFB$ is a right angle.  Thus triangles $ABE$ and $BCF$ are congruent by $HA$.  From the Pythagorean Theorem and congruence of corresponding parts we then have:
$CE^2=CF^2+EF^2=CF^2+(BE+BF)^2=BE^2+(BE+AE)^2$
Putting in $BE=6, AE=2$ then gives just $CE=10$.
A: From point $E$ bring a parallel to line segment $BC$. Extend $CD$ until it intersects it on the point $N$.
Now you have the right triangle $CNE$ on which $EC$ is the hypotenuse. Thus we only need to find $EN$ and $DN$ in order to apply the Pythagorean Theorem.
$$EC^2=EN^2+NC^2$$ 
But $NC=CD+DN=2\sqrt10+DN$ and $DN$ is equal to the height of the right triangle $AED$, so $$\frac1{DN^2}=\frac1{AE^2}+\frac1{ED^2}=\frac1{4}+\frac1{36}$$
and $EN$ is-via the Pythagorean Theorem applied on $END$
$$EN^2=6^2-DN^2$$
