7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$.
Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Let $O E=x, A E=y$. Then we arrive at the system $\left\{\begin{array}{c}x^{2}+y^{2}=2, \\ (1+\sqrt{3}-y)^{2}+x^{2}=4\end{array}\right.$. Solving the system, we get $\quad y=1, x=1$. Then $\quad \angle D A C=\angle B C A=45^{\circ}$, $B C=\sqrt{6}, \angle F C A=\operatorname{arctg} \frac{1}{\sqrt{3}}=30^{\circ}$,

$\angle A B E=30^{\circ}, \quad A B=2$.

Answer: 2.