Proving some segments equal

Question

I have the following diagram. I am not given the equality marks that are in the diagram. What I am given is that $ABC$ is a right triangle with equal legs, $AD=AB$, and that $CDE$ is a right angle. What I have to prove is those equality marks: $BE=ED=DC$. I've prove $BE=ED$ by constructing two triangle by drawing a line from $A$ to $E$ and proving $ABE$ congruent to $ADE$, but I can't take that to them equaling $DC$. It may be something trivial that I'm missing.

Answer 1

This image makes you think like it’s always true because of the way AB and BC seem to be equal, but that’s not a condition.

Sorry for the bad drawing I’m on my phone.

Edit: if there is indeed a 45 degree angle in BAC then The triangle DEC is isocèle by noticing that the sum of the angles must be 180.

Answer 2

If you take a look at the following triangle you will see that a further condition is missing in order to have $ED=EC$.

That further condition may be that $ABC$ is isosceles.

Answer 3

The proposed solution to the problem, as stated, is false.

If and only if the two marked sides are equal in length, then $\angle DCE = 45^\circ$. But the overall figure can be drawn without $\angle DCE = 45^\circ$.

Answer 4

I agree with David, it appears false. Is there something not to scale?

It's tough to tweak exactly on a mobile phone but this is how I copied and rotated the triangle.

There's a skinny rectangle that remains unfilled.