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 $\overline{EB}$: $$(\overline{CE}=\overline{CA}, \overline{CB}=\overline{CD},\angle ECB=90^0+\angle ACB=\angle ACD) \Rightarrow \triangle ECB=\triangle ACD \Rightarrow \angle CBE=\angle CDA \Rightarrow CPBD \text{ is cyclic} \Rightarrow \angle BPD = \angle BCD \Rightarrow \angle BPD = 90^0 \square.$$