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?
Continue with $BE = \sqrt{8^2+4^2} = 4\sqrt5$ and recognize that the triangle BDE is isosceles with $BD =DE=5$,

$$\cos \angle ABE =\frac{AB}{BE} = \frac1{\sqrt5}, \&gt;\&gt;\&gt;\&gt;\&gt;
\cos \angle DBE =\frac{BE/2}{BD} = \frac2{\sqrt5}
$$

Thus,

$$\angle ABD = \angle ABE + \angle DBE = \arccos \frac1{\sqrt5}
+ \arccos \frac2{\sqrt5}=90^\circ$$
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.