Let $H$ - point intersection of diagonals, $X$ - second point intersections of $(BHA), (CHD)$. Note, that F lie on $(CHD)$. From Reim theorem $X$ lie on $BF$. $XH$ is rad axe of $(BHA), (CHD)$ so it perpendicular to it line of centers(it's obvious that it is midpoints of $CD, AB$ so it is mixline of trapezoid and it is parallel to the bases) so $HX$ perpendicular to $AD$, but $AHC$ is isosceles so $HX$ is bisector of $AHC$ and $\angle(AH, HX) = \angle 45^{\circ}$. Then, $\angle ABF = \angle 45^{\circ}$. $AB//CF$ so $\angle BFC = \angle 45^{\circ}$, but $BH = HC$ and $\angle BHC = \angle 90^{\circ} = 2\angle 45^{\circ} = 2\angle BFC$ so $H$ is center of circircle of $(BCF)$ so $BH = HF$, $\angle FCD = \angle FHD = 2\angle FBD$.