​ help

ECB = 180 - 55 = 125

Draw a line from F to A   so you have two right angles in a pentagon

Interior angles of a pentagon =  (n-2) 180 = 540

then   CEF = 540 - 90 -90 -120 - 125 = 115 ^o

First, I'll just disregard the fact that this is an AoPS question.

 BCD is a straight line, so we have BCE = 180 . We draw a line CG, which is parallel to AB and EF.

AB || CG, so

\(\angle ECG = \angle BCE - \angle BCG = 125^\circ - 60^\circ = 65^\circ\)

CG and EG are parallel so

\(\angle CEF = 180^\circ - \angle ECG = \boxed{115^\circ}.\)

Since AB and EF are parallel.....drop a perpendicular  from AB to EF

Then  both of the angles formed by this perpendicular  = 90°

And angle ECD is supplemental to angle  BCE....so angle BCE  = 180 - 55   = 125°

And we  have  an irregular pentagon......and the sum of the interior  angless =540°

So  angle CEF  can be found as

CEF  +  90  +  90  +  125  + 120    = 540

CEF  + 425  = 540

CEF  = 540  - 425  =  115°