7. (4 points) Given a right triangle $A B C$, where angle $A$ is 60 degrees, and the hypotenuse $A B$ is $4+4 \sqrt{3}$. Through vertex $B$, a line $p$ parallel to $A C$ is drawn. Points $D$ and $E$ are placed on line $p$ such that $A B=B D, B C=B E$. $F$ is the intersection point of lines $A D$ and $C E$. Find what the perimeter of the triangle can be

$$
\begin{gathered}
4 \sqrt{2}+2 \sqrt{6}+18+10 \sqrt{3} \\
2 \sqrt{3}+2+2 \sqrt{2} \\
18+10 \sqrt{3}+4 \sqrt{6}+6 \sqrt{2} \\
2+2 \sqrt{3}+2 \sqrt{6}
\end{gathered}
$$
Solution: First, we calculate the sides of triangle $ABC$: side $AC$ lies opposite the angle $30^{\circ}$, so it is half the hypotenuse, which is $2+2\sqrt{3}$. Now, using the Pythagorean theorem, we calculate $BC = \sqrt{(4+4\sqrt{3})^2 - (2+2\sqrt{3})^2} = \sqrt{16+32\sqrt{3}+48-4-8\sqrt{3}-12} = \sqrt{36+24\sqrt{3}+12} = 6+2\sqrt{3}$.

Triangle $CBE$ is a right isosceles triangle with a right angle at $B$, so line $CE$ intersects $p$ at an angle of $45^{\circ}$. Triangle $ABD$ is isosceles with the angle at vertex $B$ being $120^{\circ}$ or $60^{\circ}$ depending on the position of point $D$. Therefore, line $AD$ intersects $p$ at an angle of $60^{\circ}$ or $30^{\circ}$.

Depending on the positions of points $D$ and $E$ on line $p$, triangle $DEF$ has angles of $30, 45, 105; 30, 135, 15; 60, 45, 75; 120, 45, 15$ degrees.

Consider the height $FH$ in triangle $DEF$. From the size of angle $E$, it follows that $EH = FH$.

In a right triangle with angles $60^{\circ}$ and $30^{\circ}$, the legs differ by a factor of $\sqrt{3}$, and triangle $DFH$ is precisely such a triangle. Therefore, in the first two cases, $DH = \sqrt{3} FH$, and in the remaining two cases, $DH = \frac{FH}{\sqrt{3}}$. Thus, $DE = |DH \pm EH|$ (the minus sign appears in the obtuse triangle $DEF$ with an obtuse angle at vertex $D$ or $E$), that is, when points $D$ and $E$ lie on the same side of $B$), and therefore, $DE = FH |\sqrt{3} \pm 1|$ or $DE = FH \left|1 \pm \frac{\sqrt{3}}{3}\right|$. Express $FH$ in terms of $DE$. Side $EF = \sqrt{2}$ as the hypotenuse in the right isosceles triangle $EFH$, and in the right triangle $DFH$, the hypotenuse is twice the shorter leg.

Now, let's calculate the desired perimeter in each case separately.

Angles $30, 45, 105$ degrees. Points $D$ and $E$ lie on opposite sides of point $B$ and $DE = BD + BE = AB + BC = 6\sqrt{3} + 10$. $FH = \frac{DE}{\sqrt{3} + 1} = \frac{6\sqrt{3} + 10}{\sqrt{3} + 1} = \frac{1}{2}(6\sqrt{3} + 10)(\sqrt{3} - 1) = 4 + 2\sqrt{3}$, then $EF = 4\sqrt{2} + 2\sqrt{6}$ and $DF = 2FH = 8 + 4\sqrt{3}$. We get a perimeter of $4\sqrt{2} + 2\sqrt{6} + 18 + 10\sqrt{3}$.

Angles $30, 135, 15$ degrees. Points $D$ and $E$ lie on the same side of point $B$ and $DE = BD - BE = AB - BC = 2\sqrt{3} - 2$. $FH = \frac{DE}{\sqrt{3} - 1} = 2$, then $EF = 2\sqrt{2}$ and $DF = 2FH = 4$. We get a perimeter of $2\sqrt{3} + 2 + 2\sqrt{2}$.

Angles $60, 45, 75$ degrees. Points $D$ and $E$ lie on opposite sides of point $B$ and $DE = BD + BE = AB + BC = 6\sqrt{3} + 10$. $FH = \frac{DE}{\frac{\sqrt{3}}{3} + 1} = \sqrt{3} \cdot \frac{DE}{\sqrt{3} + 1} = 4\sqrt{3} + 6$, then $EF = 4\sqrt{6} + 6\sqrt{2}$ and $DF = \frac{2}{\sqrt{3}} FH = 8 + 4\sqrt{3}$. We get a perimeter of $18 + 10\sqrt{3} + 4\sqrt{6} + 6\sqrt{2}$.

Angles $120, 45, 15$ degrees. Points $D$ and $E$ lie on the same side of point $B$ and $DE = BD - BE = AB - BC = 2\sqrt{3} - 2$. $FH = \frac{DE}{1 - \frac{\sqrt{3}}{3}} = \sqrt{3} \cdot \frac{DE}{\sqrt{3} - 1} = 2\sqrt{3}$, then $EF = 2\sqrt{6}$, and $DF = \frac{2}{\sqrt{3}} FH = 4$. We get a perimeter of $2 + 2\sqrt{3} + 2\sqrt{6}$.