How to show that these two lines are perpendicular? Let $\triangle AEE'$ be an isosceles triangle with $\angle EAE'=90^\circ$ such that $AE=AE'$ and such that $A$, $E$ and $E'$ lie on the circle $c_1$. Let $\triangle ADD'$ be an isosceles triangle with $\angle DAD'=90^\circ$ such that $AD=AD'$ and such that $A$, $D$ and $D'$ lie on the circle $c_2$. Let $S$ be the intersection of the lines $DE'$ and $DE'$.
The picture below summarizes the situation: 
I want to prove that $D'E \perp DE'$, i.e. $\angle ESE'=90^\circ$.
I have already shown that $\triangle AED'$ and $\triangle AE'D$ are congruent. How do I continue from here? I hope that it is possible to avoid using a coordinate system.
Thanks in advance!
 A: Here is another way to prove without using the fact that $S$ lies on intersection of circles :
$$ \begin{align} &\angle ADD' =  \angle AD'D = 45^{\circ} \\~\\ \end{align} $$ 
$$\begin{align}& \overline{AE} \cong \overline{AE'} \\ &\angle EAD' \cong \angle E'AD \\
&\overline{AD'} \cong \overline{AD} \\ &\implies \triangle AED' \cong \triangle AE'D ~~\text{By SAS}\\&
\angle ADE' \cong \angle AD'E ~\color{gray}{\text{By CPCTC}} \\
 \end{align}$$

$$\begin{align}&\implies \angle DD'S  = 45+x ~~\text{and}~~\angle D'DS = 45-x \\~\\ &\implies \angle DSD' = 90 \\&~~\color{gray}{\text{By triangle sum property in }  \triangle DSD'
}\end{align} $$
A: Here's a terse proof with complex numbers. Take $A$ to be the origin and let the coordinates of $D,E,D',E'$ be $iz,w,z,i w$ respectively. Then be construction we have isosceles triangles formed by the points $\{0,z,iz\}$ and $\{0,w,iw\}$. The angle formed by $DE'$ and $D'E$ is then found as the argument of $\dfrac{z-w}{i(z-w)}=-i$, and is thus $90^\circ$.
A: If you have already shown that $AED'$ and $AE'D$ are congruent, then observe that you can pivot the first triangle about $A$ to get the second. The angle of rotation is $\angle EAE'$, or $90^\circ$, because the edge $AE$ is sent to $AE'$.
Therefore every corresponding pair of edges between the two triangles is perpendicular, including in particular $DE'$ and $D'E$.
A: Let $P$ be the intersection of $DS$ and $AD'.$
Then $\angle PD'S = \angle AD'E= \angle ADE' = \angle ADP,$
using the fact that $\triangle AD'E \cong \triangle ADE'$. 
Also $\angle APD=\angle SPD'$ since they are a pair of vertical angles.
Two angles of $\triangle APD$ are congruent to corresponding angles in $\triangle SPD'$,
so the two triangles are similar, hence $\angle D'SP=\angle PAD = 90^\circ$,
hence  $D'E \perp DE'.$
