I'm Stumped

Extend CE and extend AB, let us call the point of intersection $L$

 - Because of vertical angles angle $LCB$ is 55 degrees.

 - Angle $CBL$ is 180 - 120 = 60 degrees.

Since all angles in a triangle add up to 180 degrees, angle $CLB$ = 65 degrees. Because of alternate interior angles, the supplement of CLB is congruent to CEF.

Therefore, CEF = 180 - 65 = $\boxed{115}$ degrees.

I went at it a liffle differently. 

Since AB & EF are parallel, draw a line between them that is perpendicular to both.

Now you have a pentagram.  The total of the interior angles of a pentogram is (5 – 2)(180) = 540

One of those right angles is                                              90 

The other right angle is                                                     90

You are given that angle ABC is                                      120

Angle ECD is 55 therefore angle ECB is (180 – 55)        125

The total of those four angles is                                       425 

Subtract the total of the four angles from the total

of the interior angles to get the fifth angle (540 – 425)     115 

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