The figure below is not drawn to scale. ABEF is a parallelogram and BCDE is a trapezium. BE//CD, AB//FD and BC//AD. DF is a straight line. $\angle ADF = 42^\circ$ and $\angle DAF = 68^\circ$.

1. Find the angle corresponding to $\angle DAF$ on the line $BE$.

Let $AD$ intersect $BE$ at a point $P$. Since $ABEF$ is a parallelogram, $AF \parallel BE$.
Using $AD$ as a transversal, the corresponding angle $\angle DPE = \angle DAF = 68^\circ$.

2. Find $\angle CBE$ using the parallel lines $BC$ and $AD$.

We are given that $BC \parallel AD$. Using $BE$ as a transversal, the corresponding angle $\angle CBE = \angle DPE = 68^\circ$.

3. Calculate $\angle BCD$ using the interior angles of the trapezium.

In trapezium $BCDE$, we are given $BE \parallel CD$.
Interior angles between parallel lines sum to $180^\circ$, so:
$\angle BCD + \angle CBE = 180^\circ$
$\angle BCD = 180^\circ - 68^\circ = 112^\circ$

1. Find $\angle AFD$ using the sum of angles in $\triangle ADF$.

In $\triangle ADF$,
$\angle AFD = 180^\circ - 68^\circ - 42^\circ = 70^\circ$ (angle sum of triangle)

2. Find $\angle ABE$ using parallel lines.

Since $AB \parallel FD$, the interior angles $\angle BAF$ and $\angle AFD$ are supplementary:
$\angle BAF = 180^\circ - 70^\circ = 110^\circ$

Since $ABEF$ is a parallelogram, $AF \parallel BE$. The interior angles $\angle BAF$ and $\angle ABE$ are supplementary:
$\angle ABE = 180^\circ - 110^\circ = 70^\circ$

(Alternatively, $\angle ABE = \angle AFD = 70^\circ$ directly since both are supplementary to $\angle BAF$)

1. Find $\angle AFD$ using the angle sum of $\triangle ADF$.

In $\triangle ADF$,
$\angle AFD = 180^\circ - 68^\circ - 42^\circ = 70^\circ$

2. Find $\angle BAF$ using the parallel lines $AB$ and $FD$.

Since $AB \parallel FD$, the interior angles are supplementary:
$\angle BAF = 180^\circ - \angle AFD = 180^\circ - 70^\circ = 110^\circ$

3. Find $\angle BEF$ using the properties of parallelogram $ABEF$.

In parallelogram $ABEF$, opposite angles are equal:
$\angle BEF = \angle BAF = 110^\circ$