4. (3 points) Given a triangle $A B C$ with angle $C=120^{\circ}$. Point $D$ is the foot of the perpendicular dropped from point $C$ to side $A B$; points $E$ and $F$ are the feet of the perpendiculars dropped from point $D$ to sides $A C$ and $B C$ respectively. Find the perimeter of triangle $A B C$, if it is known that triangle $E F C$ is isosceles and its area is $\sqrt{3}$.
Answer: $16+8 \sqrt{3}$

## Solution:

Triangle $E F C$ is isosceles. Since $\angle C$ in it is obtuse, point $C$ is its vertex, which means $E C=$ $F C$. But then the right triangles $E C D$ and $F C D$ are equal by the leg and hypotenuse, so $E D=$ $F D$, that is, point $D$ lies on the bisector of angle $C$, from which triangle $A B C$ is also isosceles.

Thus, triangles $A B C$ and $E F C$ are similar. Let's find the similarity coefficient. First, $C E: C D=\cos \angle A C D=\frac{1}{2}$. Second, $C D: A C=\cos \angle A C D=\frac{1}{2}$. Therefore, $C E: A C=\frac{1}{4}$.

Next, $\sqrt{3}=S_{E F C}=\frac{E C \cdot F C \cdot \sin \angle A C D}{2}=\frac{E C^{2} \sqrt{3}}{4}$, from which $E C=2$. Hence, $E F=2 E C \sin 60^{\circ}=2 \sqrt{3}$. Therefore, $P_{E F C}=4+2 \sqrt{3}, P_{A B C}=4 P_{E F C}=16+8 \sqrt{3}$.