In the figure below, ABCD is a rectangle. The points E and F lie on the rectangle ABCD. $\angle DFE = 105^\circ$, $\angle EAC = 32^\circ$ and $\angle BCE = 50^\circ$.

1. Find $\angle BCA$ using the right-angled triangle $ABC$.

Since $E$ lies on the line segment $AB$, $\angle BAC = \angle EAC = 32^\circ$.
In the rectangle $ABCD$, $\angle ABC = 90^\circ$.
Considering $\triangle ABC$:
$\angle BCA = 180^\circ - 90^\circ - 32^\circ = 58^\circ$

2. Calculate $\angle ECA$.

From the figure, $\angle BCA = \angle BCE + \angle ECA$.
$\angle ECA = \angle BCA - \angle BCE$
$\angle ECA = 58^\circ - 50^\circ = 8^\circ$

1. Find $\angle CEB$ using the right-angled triangle $BCE$.

In rectangle $ABCD$, $\angle B = 90^\circ$.
In $\triangle BCE$, the sum of angles is $180^\circ$.
$\angle CEB = 180^\circ - 90^\circ - 50^\circ = 40^\circ$

2. Use alternate interior angles to find $\angle BEF$ and then $\angle FEC$.

Since $AB \parallel DC$, alternate interior angles are equal.
$\angle BEF = \angle DFE = 105^\circ$
From the figure, $\angle BEF = \angle FEC + \angle CEB$.
$105^\circ = \angle FEC + 40^\circ$
$\angle FEC = 105^\circ - 40^\circ = 65^\circ$