What is the value of $\angle ABD$ in the following figure? 

In the given figure $\angle BAE, \angle BCD$ and $\angle CDE$ are
  right angles and $AB = 4, BC=3, CD=4$  and $DE=5$. What is the value
  of $\angle ABD$?

I found this problem in a sheet of contest math problems. However the problem stated in the sheet was "Find the value of $AE$ ". I proved the problem using $BE^2 = 8^2+4^2$. Then I noticed that it could be proved by making rectangle $ABED'$. I tried to prove why $\angle ABD$ is a right angle only with the given conditions but failed. Is the $\angle ABD$ even a right angle (then how to prove it) or it can have different values? 
 A: Continue with $BE = \sqrt{8^2+4^2} = 4\sqrt5$ and recognize that the triangle BDE is isosceles with $BD =DE=5$,
$$\cos \angle ABE =\frac{AB}{BE} = \frac1{\sqrt5}, \>\>\>\>\>
\cos \angle DBE =\frac{BE/2}{BD} = \frac2{\sqrt5}
$$
Thus,
$$\angle ABD = \angle ABE + \angle DBE = \arccos \frac1{\sqrt5}
+ \arccos \frac2{\sqrt5}=90^\circ$$
A: The original question is under-defined. 
E.g. We could have $ACB$ as a straight line, with the desired angles still right.
Explicitly, $ A = (0,0), B = (0, -4), C = (0, -1), D = (4, -1), E = (4, 4) $ which gives us $ AE = 4\sqrt{2}$.   

What we have is $AE = 8 \Leftrightarrow \angle ABD = 90^ \circ$.    
As Quanto's solution indicates: If $ AE = 8$, then $\angle ABD = 90^ \circ$. 
If $ \angle ABD = 90^ \circ$, then we can easily show $AE = 8$ (E.g. by coordinate geometry, or length chasing.)
A: BD = 5.
Drawing a perpendicular from D to AE at G,
DG = 4 so EG = 3
and AG = 5 so AE = 8.
As to the angles,
EDG = CDB (both 3-4-5)
so CDE = AED = CBD
so BDG = 90
so ABD = 90.
