ABCD 是一个正方形，CDEF 是一个菱形。EFG 是一个三角形。ECG 是一条直线。(\angle EFG = 88^\circ)。

1. Identify the isosceles triangle formed by the sides of the rhombus

In the rhombus $CDEF$, all four sides are equal in length. Therefore, $CD = DE$, which makes $\triangle CDE$ an isosceles triangle.

2. Calculate the required angle using the angle sum of the triangle

Since $\triangle CDE$ is isosceles with $CD = DE$, the base angles are equal: $\angle DCE = \angle DEC = 67^\circ$.
Using the angle sum of a triangle:
$\angle EDC = 180^\circ - 67^\circ - 67^\circ = 46^\circ$

1. Find $\angle FEC$ using the properties of a rhombus.

In the rhombus $CDEF$, the diagonal $CE$ bisects the interior angle $\angle DEF$. Therefore, $\angle FEC = \angle DEC = 67^\circ$.
Since $ECG$ is a straight line, the angle $\angle FEG$ is the same as $\angle FEC$, so $\angle FEG = 67^\circ$.

2. Calculate $\angle EGF$ using the angle sum of $\triangle EFG$.

In $\triangle EFG$, the sum of the interior angles is $180^\circ$.
$\angle EGF = 180^\circ - \angle FEG - \angle EFG$
$\angle EGF = 180^\circ - 67^\circ - 88^\circ = 25^\circ$

1. Find the interior angle $\angle DCF$ of the rhombus.

In rhombus $CDEF$, $CD = DE$, so $\triangle CDE$ is isosceles.
$\angle DCE = \angle DEC = 67^\circ$
$\angle CDE = 180^\circ - 2(67^\circ) = 46^\circ$
Since consecutive angles in a rhombus are supplementary:
$\angle DCF = 180^\circ - 46^\circ = 134^\circ$

2. Find $\angle BCF$ using angles at a point.

The angles around point $C$ are $\angle BCD$ (interior of square), $\angle DCF$ (interior of rhombus), and $\angle BCF$.
$\angle BCD = 90^\circ$
$\angle BCF = 360^\circ - 90^\circ - 134^\circ = 136^\circ$

3. Use the isosceles triangle $\triangle BCF$ to find $\angle CBF$.

Since $ABCD$ is a square and $CDEF$ is a rhombus sharing side $CD$, we have $BC = CD = CF$.
Therefore, $\triangle BCF$ is an isosceles triangle with $BC = CF$.
$\angle CBF = \frac{180^\circ - 136^\circ}{2} = \frac{44^\circ}{2} = 22^\circ$