## Area of a triangle

In square $ABCD$, $E$ is the midpoint of side $\overline{AB}$ and $F$ is a point of side $\overline{AD}$ such that $F$ is twice as near from $D$ as from $A$. $G$ is the intersection of the line segments $\overline{DE}$ and $\overline{CF}$. If $AB = 1\text{ cm}$, find the area of $\triangle CDG$.
There are several ways to solve this problem, please show your solutions.

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