How can one prove that (AD) and (EB) are orthogonal? ABC is a triangle, we make outside two squares CBGD and ACEH.
I have to show that (AD) and (EB) are orthogonal. So I'm sure that there is many ways to solve this exercise. 
I did it using the scalar product:
To prove that (AD) and (EB) are orthogonal we have to show that
$\vec{AD}\cdot\vec{EB}=0$.
Using theorem of shasles:
$$\vec{AD}\cdot\vec{EB}=(\vec{CD}-\vec{CA})\cdot(\vec{CB}-\vec{CE})$$
We have $\vec{CD}\cdot\vec{CB}=0$, $\vec{CA}\cdot\vec{CE}=0$ and $\vec{CA}\cdot\vec{CB}=-\vec{CD}\cdot\vec{CE}$. So using the scalar product we can prove that (AD) and (EB) are orthogonal.
But is there any other solutions with other ways? 

Let $P$ be the intersection point of $\overline{AD}$ and $\overline{EB}$:
$$(\overline{CE}=\overline{CA}, \overline{CB}=\overline{CD},\angle ECB=90^0+\angle ACB=\angle ACD) \Rightarrow \triangle ECB=\triangle ACD \Rightarrow \angle CBE=\angle CDA \Rightarrow CPBD \text{ is cyclic} \Rightarrow \angle BPD = \angle BCD \Rightarrow \angle BPD = 90^0 \square.$$
