Denote $E$ as the intersection point of $CF$ and $BG$ and $O$ as the center of $k$. We need to prove that $\angle FEG = \frac{1}{2} \angle FOG$. Since $BC$ is $1$ cm long, and legs $AB$ and $AC$ are $2$ cm long, we have: $BC = 1/2AC = 1/2AB = AF = BF = CG = DG$. Also, $FG // BC$ since it is a medial line of $ABC$. Thus, $\angle BCF = \angle CBG = \angle BFC = \angle CBG = \angle CFG = \angle BGF$. Notice that since $O$ is the center of $k$, which is tangent to $AB$ and $AC$, $OF \perp AB$ and $OG \perp AC$. Thus, $A, F, O, G$ are concyclic, which means $\angle AOF = \angle AGF = \angle AFG = \angle AOG$, and: $\angle FEG = 180^o - 2\angle MGF = \angle AGF = \angle AOF = \frac{1}{2}\angle FOG$, which is the desired conclusion.

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