ABCD is a rectangle. AE and DG are straight lines. $\angle ADF = \angle FDE$.

1. Use the property of a rectangle to set up an equation for the angles at vertex D.

Since $ABCD$ is a rectangle, the interior angle at $D$ is $90^\circ$. Thus, $\angle CDA = 90^\circ$.
The rays from $DC$ to $DA$ are $DC$, $DE$, $DF$ (which is the diagonal $DB$), and $DA$.
Therefore, $\angle CDE + \angle FDE + \angle ADF = 90^\circ$.

2. Solve for $\angle FDE$ using the given angle equalities.

We are given $\angle CDE = 12^\circ$ and $\angle ADF = \angle FDE$.
Substituting these into the equation:
$12^\circ + \angle FDE + \angle FDE = 90^\circ$
$2\angle FDE = 90^\circ - 12^\circ = 78^\circ$
$\angle FDE = \frac{78^\circ}{2} = 39^\circ$

1. Find $\angle ADE$

Since $ABCD$ is a rectangle, $\angle ADC = 90^\circ$.
$\angle ADE = 90^\circ - \angle CDE = 90^\circ - 12^\circ = 78^\circ$.

2. Find $\angle DAE$ using the sum of angles in $\triangle ADE$

In $\triangle ADE$, the sum of angles is $180^\circ$. Since $F$ lies on $AE$, $\angle AED = \angle DEF = 43^\circ$.
$\angle DAE = 180^\circ - \angle ADE - \angle AED = 180^\circ - 78^\circ - 43^\circ = 59^\circ$.

3. Find $\angle FAG$

Since $\angle DAB = 90^\circ$ and $G$ lies on $AB$,
$\angle FAG = \angle EAB = 90^\circ - \angle DAE = 90^\circ - 59^\circ = 31^\circ$.