Right Triangle Angle problem

Question

I know that this can be easy, but I found it a bit difficult so maybe someone can just explain or give a hint to me.

I have a right triangle $ABC$ with points $D$ and $E$ on its hypotenuse so $AB=AE$ and $BC=CD$. I need to find angle $DBE$.

So here's two isosceles triangles (or isn't it?) $BDC$ and $ABE$. I think I need these to solve that angle. But what to do next?

Answer 1

Hint:

Let $\angle A = \alpha$ so that $\angle C=90°-\alpha$.

From the isosceles triangles you have: $$ \angle (ABE)=\frac{180°-\alpha}{2} \qquad \angle(DBC)=\frac{90°+\alpha}{2} $$ now use: $$ 90°=\angle(ABC)=\angle (ABE)+\angle(DBC)- \angle(DBE) $$