Trianglogram!

In ||gm ABCD

∠ACD=∠CAB= 42°

∠ABC= 180-114= 66°

Now, ∠CAB+∠ABC= 42+66 =108° = y

Since BC=CF, ∠CBF=∠CFB= x

In triangle CBF

y + x + x = 180°

108 + 2x = 180°

2x = 72°

x = 36°

Therefore, x + y = 108+36 = 144°

Since the sum of angles in a triangle is $180^\circ$ and $\overline{BC} = \overline{CF},$ $y + 2x = 180^\circ \Rightarrow x = \frac{180^\circ - y}{2}.$ Also $\angle DCB = \angle EBC = 114^\circ$ by the transversal theorem, and $\angle ACB = \angle DCB - 42^\circ = 72^\circ.$

Angle $y$ is supplementary to $\angle ACB$ so its measure is therefore $y = 180^\circ - 72^\circ = 108^\circ.$ Now from the equation above $x + y = \frac{180^\circ - y}{2} + y \\ = \frac{180^\circ + y}{2} \\ = \frac{288^\circ}{2} \\ = \boxed{144^\circ}$