5. Given in the Cartesian coordinate system $x O y$, the three vertices of $\triangle A B C$ are $A(2 \sqrt{2}, \sqrt{2}+\sqrt{6})$, $B(\sqrt{2}$, $\sqrt{2})$, and $C(5 \sqrt{2}, \sqrt{2})$. Then the coordinates of the intersection point of the altitude from $A$ to side $B C$ and the angle bisector of $\angle A B C$ are $\qquad$
5. $\left(2 \sqrt{2}, \sqrt{2}+\frac{\sqrt{6}}{3}\right)$.

Let the altitude from $A$ to $BC$ and the angle bisector of $\angle ABC$ intersect at point $P(2 \sqrt{2}, \sqrt{2}+h)$. Then $\tan \angle ABC=\frac{\sqrt{6}}{\sqrt{2}}, \tan \angle PBC=\frac{h}{\sqrt{2}}$. Since $\angle ABC=2 \angle PBC$, we have $\frac{\sqrt{6}}{\sqrt{2}}=\frac{2 \frac{h}{\sqrt{2}}}{1-\left(\frac{h}{\sqrt{2}}\right)^{2}} \Rightarrow h=\frac{\sqrt{6}}{3}$.