Angle chasing with two tangent circles. The bigger circle $\Omega$ is tangent to the smaller circle $\omega$.
Also, $GE=2CG$.
We have to find $\angle DEC$.

MY WORK SO FAR.
I proved using the Alternate Segment Theorem that:
$$GF\parallel ED$$
And that,
$$\angle DCH=\angle HCE=45°$$
Also,
$$GF=GH$$

The fact that there are points $G$ and $E$, $G$ on the small circle, and $E$ on the big circle such that $\vec{CE}=3 \vec{CG}$ means that the circles are homothetic with the homothety with center $C$ and scale factor 3 (http://www.cut-the-knot.org/Curriculum/Geometry/Homothety.shtml). Thus, we also have, for the centers, $\vec{CA}=3 \vec{CB}$.
Therefore, we can assume, WLOG, that the circles are resp. with centers $(1,0)$ and $(3,0)$, and radii 1 and 3. Let $ED$ the tangent issued from $E$ and tangent in $H$ to the small circle.
With the notations of the figure below:
$$\sin(2\alpha)=\dfrac{BH}{BA}=\dfrac{1}{2}$$
Thus, assuming angle $2\alpha < tfrac{\pi}{2}$, we deduce that  $2\alpha=\dfrac{\pi}{6}$.
Therefore, by central angle theorem (see (https://www.geogebra.org/m/eNA87edZ)):

$$\alpha=angle(DEC)=\dfrac{\pi}{12}$$
  

