Square Geometry Problem

Question

https://i.stack.imgur.com/K0rV1.jpg

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.

Answer 1

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 $EC = 10$.

Answer 2

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 $END$

$$EN^2=6^2-DN^2$$

Answer 3

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 $CE=10$.