Squares constructed externally on the sides of a triangle and concurrent lines On the sides $BC, CA$ and $AB$ of the triangle $ABC$ we construct externally the squares $BCDE, ACFG $ and $ABHI$. Denote $A', B'$ and $C'$ the intersectiond points of the lines $BF$ and $CH$, $AD$ and $CI$, respectively $AE$ and $BG$. Prove, that $AA', BB'$ and $CC'$ are concurrent.

Any nice idea? Thank you!

The lines $AA',BB',CC'$ are the altitudes of the triangle, so they intersect in the orthocenter.
To show that these are in fact altitudes, you can use a computation on coordinates. W.l.o.g. assume the following coordinates:
$$A=\begin{pmatrix}-1\\0\end{pmatrix}\qquad
B=\begin{pmatrix}1\\0\end{pmatrix}\qquad
C=\begin{pmatrix}x_C\\y_C\end{pmatrix}$$
From these you get
$$
G = A+\begin{pmatrix}0&-1\\1&0\end{pmatrix}(C-A) =
\begin{pmatrix}-1-y_C\\1+x_C\end{pmatrix} \\
E = B+\begin{pmatrix}0&1\\-1&0\end{pmatrix}(C-B) =
\begin{pmatrix}1+y_C\\1-x_C\end{pmatrix} \\
AE: (x_C-1)x+(y_C+2)y=1-x_C \\
BG: (x_C+1)x+(y_C+2)y=1+x_C \\
C' = \begin{pmatrix}x_C\\\tfrac{1-x_C^2}{2+y_C}\end{pmatrix}
$$
so you see that $C$ and $C'$ have the same $x$ coordinate, therefore $CC'\perp AB$. The other orthogonalities follow from the symmetry of the construction.
The lines joining the points as well as their intersection were actually computed using cross products of homogeneous coordinates:
$$
g_{AE} = A_h\times E_h =
\begin{pmatrix}-1\\0\\1\end{pmatrix}\times
\begin{pmatrix}1+y_C\\1-x_C\\1\end{pmatrix}=
\begin{pmatrix}x_C-1\\y_C+2\\x_C-1\end{pmatrix} \\
g_{BG} = G_h\times B_h =
\begin{pmatrix}-1-y_C\\1+x_C\\1\end{pmatrix}\times
\begin{pmatrix}1\\0\\1\end{pmatrix}=
\begin{pmatrix}x_C+1\\y_C+2\\-x_C-1\end{pmatrix} \\
C'_h = g_{AE}\times g_{BG} =
\begin{pmatrix}x_C-1\\y_C+2\\x_C-1\end{pmatrix}\times
\begin{pmatrix}x_C+1\\y_C+2\\-x_C-1\end{pmatrix}=
\begin{pmatrix}-2x_Cy_C-4x_C\\2x_C^2-2\\-2y_C-4\end{pmatrix}
$$
