I'm bored.

At first we can find out that $\angle ABF =60^{o}$ in many ways, and this means that $\overline{AB} \cong \overline{AC}$ (As result of its similarity with any half equilateral triangle). Then, we know that $\bigtriangleup ABC$ is isosceles $\Rightarrow \angle DBC=15^{o}$ . As the line segment $\overline{FC}$ is an altitude and at once a median, we can notice that $\bigtriangleup DBC$ is, too, isosceles, so we have: $(\overline{BC} \cong \overline{DC} \wedge \angle EDC=75^{o}) \Rightarrow \bigtriangleup EDC \cong \bigtriangleup ABC \Rightarrow \angle DEC=30^{o}$ And, by using the exterior angle theorem in tiangles $\bigtriangleup EDC$ and $\bigtriangleup DBC$ we find: $x=90^{o}+45^{o}=135^{o}$