ABCD 是一个菱形，CDEF 是一个平行四边形。
求 $\angle ACG$。

1. Find the angle $\angle ACD$ in the rhombus $ABCD$.

In rhombus $ABCD$, the adjacent angles sum to $180^\circ$, so $\angle BCD = 180^\circ - 46^\circ = 134^\circ$.
The diagonal $AC$ bisects $\angle BCD$, so:
$\angle ACD = \frac{134^\circ}{2} = 67^\circ$

2. Find the angle $\angle FCD$ in the parallelogram $CDEF$.

In parallelogram $CDEF$, adjacent angles sum to $180^\circ$. Assuming the given angle $128^\circ$ refers to the interior angle $\angle EFC$ (or $\angle CDE$), the adjacent angle $\angle FCD$ is:
$\angle FCD = 180^\circ - 128^\circ = 52^\circ$

3. Calculate $\angle ACG$.

Since $F, G, C$ are collinear and $G$ lies on $AD$, the angle $\angle GCD$ is the same as $\angle FCD = 52^\circ$.
From the figure, $\angle ACD = \angle ACG + \angle GCD$.
$\angle ACG = \angle ACD - \angle GCD = 67^\circ - 52^\circ = 15^\circ$