Proving $DE \perp BC$ in a circle by knowing the diameter I have the following circle:

I know that $AB$ is the diameter of the circle.
$C$, $D$ and $E$ are all located on the circle so the arc $AE$ is equal to $DC$. I'm trying to prove that $DE \perp BC$
I know that $AB$ is diamter so $\measuredangle C=90^{\circ}$. Also from the the arcs I can understand that $\measuredangle ABE = \measuredangle CBD$ but now I'm stuck.
 A: Connect $A$ and $D$.  We know that arc $AE$ is congruent to arc $DC$, so it follows that $\angle DAC\cong\angle ADE$.  Since those are alternate interior angles of $\overline{AC}$ and $\overline{DE}$ cut by $\overline{AD}$, we can conclude that $\overline{AC}\parallel\overline{DE}$.  Finally, now considering $\overline{CB}$ as a transversal and the fact that $m\angle ACB=90^\circ$, we may conclude that $\overline{DE}\perp\overline{BC}$.
A: Let $x=∠DBC, y=∠CBA, z=∠BAD$
$ΔBDA$ is a right triangle, we get $x+y+z=90°$. 
Arc AE = Arc DC  $\;→ ∠ABE = x$
Arc BD  $\quad\quad\quad\quad → ∠BED = z$ 
Let F be intersection of BC and DE. For $ΔBFE$:
$$∠BFE = 180° - (∠FBE + ∠BEF)= 180° - ((x+y) + z) = 90°$$
A: 
Because of same arcs AE and CD and shared arc AC, we have 
$$\angle AED = \angle CDE\tag{1}$$ 
Also, because of cyclic quadrilateral ACDE. we have 
$$\angle AED + \angle DCA = 180^\circ\tag{2}$$ 
Comebine (1) and (2), we have
$$\angle DCA + \angle CDE = 180^\circ$$
which means, 
$$AC\> || \>DE$$
Since $AC ⊥ CB$ because AB is a diameter and $AC\> || \>DE$, we conclude
$$DE ⊥ CB$$
