How to show that these two lines are perpendicular?

Question

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!

Answer 1

Here is another way to prove without using the fact that $S$ lies on intersection of circles :

$$ \begin{align} &\angle ADD' = \angle AD'D = 45^{\circ} \\~\\ \end{align} $$ $$\begin{align}& \overline{AE} \cong \overline{AE'} \\ &\angle EAD' \cong \angle E'AD \\ &\overline{AD'} \cong \overline{AD} \\ &\implies \triangle AED' \cong \triangle AE'D ~~\text{By SAS}\\& \angle ADE' \cong \angle AD'E ~\color{gray}{\text{By CPCTC}} \\ \end{align}$$

$$\begin{align}&\implies \angle DD'S = 45+x ~~\text{and}~~\angle D'DS = 45-x \\~\\ &\implies \angle DSD' = 90 \\&~~\color{gray}{\text{By triangle sum property in } \triangle DSD' }\end{align} $$

Answer 2

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 $90^\circ$.

Answer 3

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 $D'E \perp DE'.$