In the figure below, ABCF is a trapezium and CDEF is a parallelogram. AF // BC, $\angle\text{EDC} = 110^\circ$ and $\angle\text{BCF} = 52^\circ$. Find $\angle\text{AFE}$.

1. Find $\angle\text{AFC}$ using the parallel lines of the trapezium.

Since $ABCF$ is a trapezium with $AF \parallel BC$, the interior angles on the same side of the transversal $CF$ are supplementary.
$\angle\text{AFC} + \angle\text{BCF} = 180^\circ$
$\angle\text{AFC} = 180^\circ - 52^\circ = 128^\circ$

2. Find $\angle\text{EFC}$ using the properties of the parallelogram.

In parallelogram $CDEF$, opposite angles are equal.
$\angle\text{EFC} = \angle\text{EDC} = 110^\circ$

3. Calculate $\angle\text{AFE}$ using angles at a point.

The angles around point $F$ sum to $360^\circ$.
$\angle\text{AFE} + \angle\text{AFC} + \angle\text{EFC} = 360^\circ$
$\angle\text{AFE} + 128^\circ + 110^\circ = 360^\circ$
$\angle\text{AFE} = 360^\circ - 238^\circ = 122^\circ$