Four points $$A, B, C$$ and $$D$$ lie on a straight line, in this order. A point $$E$$, not on the line, satisfies $$\angle AEB = \angle BEC = \angle CED = 45^\circ$$. Let $$F$$ and $$G$$ be the midpoints of $$AC$$ and $$BD$$, respectively. If $$\angle FEG = x^\circ$$, what is the value of $$x$$?

Since $$\angle AEC = \angle BED = 90^\circ$$, the midpoint-of-hypotenuse property gives $$FA = FE = FC$$ and $$GB = GE = GD$$. Writing $$\angle EAC = \alpha$$, the isosceles triangle $$AFE$$ gives $$\angle EFG = 2\alpha$$, while the isosceles triangle $$BGE$$ gives $$\angle EGF = 180^\circ - 2(45^\circ + \alpha) = 90^\circ - 2\alpha$$. The angle sum in triangle $$EFG$$ then yields $$\angle FEG = 180^\circ - 2\alpha - (90^\circ - 2\alpha) = 90^\circ$$, so $$x = 90$$.