Collapsing Angle

As $G$ is midpoint of $\overline{FE}$ , the diagonal $\overline{AC}$ passes through $G$ . As $\angle DAC= \angle CEF =45^\circ$ the quadrilateral $ADEG$ is cyclic, so $\angle DAE =x \Rightarrow \tan(x)=0.5$

Draw CG and note that triangle CGE is an isosceles right triangle with EG = GC. Let H be the midpoint of GC. Since E is the midpoint of DC, we have EH parallel to DG. Then angle HEG = angle EGD = x. Therefore tan x = tan angle HEG = HG/EG = 1/2.

Here is another approach. Drop a perpendicular from G to the midpoint, H, of EC.The angle DGH has a tangent of (3/4)/(1/4) = 3. The angle DGH = x + 45. The tan(x + 45) = (tan(x) + tan(45)) / (1 - tan(x)tan(45)) => 3 = (tan(x) + 1) / (1 - tan(x)). Collect terms, tan(x) = 1/2.