In the figure below, ABCG is a parallelogram and CDEF is a trapezium. EF = ED and BCD is a straight line. $\angle AGC = 113^\circ$, $\angle GFC = 115^\circ$, $\angle EFD = 56^\circ$ and $\angle CDF = 48^\circ$.

1. Find $\angle EDF$ using the isosceles triangle $\triangle EFD$.

Since $EF = ED$, $\triangle EFD$ is an isosceles triangle.
Therefore, $\angle EDF = \angle EFD = 56^\circ$.

2. Identify the parallel sides of the trapezium CDEF.

In trapezium $CDEF$, either $CD \parallel FE$ or $CF \parallel DE$.
If $CD \parallel FE$, then alternate angles $\angle CDF$ and $\angle EFD$ would be equal. However, $\angle CDF = 48^\circ$ and $\angle EFD = 56^\circ$, so $CD$ is not parallel to $FE$.
Thus, $CF \parallel DE$.

3. Find $\angle CFD$ using alternate angles.

Since $CF \parallel DE$, alternate angles are equal:
$\angle CFD = \angle EDF = 56^\circ$.

4. Calculate $\angle FCD$ using the sum of angles in $\triangle FCD$.

In $\triangle FCD$, the sum of angles is $180^\circ$.
$\angle FCD = 180^\circ - \angle CFD - \angle CDF$
$\angle FCD = 180^\circ - 56^\circ - 48^\circ = 76^\circ$.

1. Find $\angle GCD$ using the properties of the parallelogram

Since $ABCG$ is a parallelogram, $AG \parallel BC$. Because $BCD$ is a straight line, $AG \parallel CD$.
Therefore, alternate angles are equal:
$\angle GCD = \angle AGC = 113^\circ$

2. Find $\angle GCF$ using the result from part (a)

From part (a), we know $\angle FCD = 76^\circ$.
$\angle GCF = \angle GCD - \angle FCD$
$\angle GCF = 113^\circ - 76^\circ = 37^\circ$

3. Calculate $\angle CGF$ using the angle sum of $\triangle CGF$

In $\triangle CGF$, the sum of angles is $180^\circ$.
$\angle CGF = 180^\circ - \angle GFC - \angle GCF$
$\angle CGF = 180^\circ - 115^\circ - 37^\circ = 28^\circ$