Let DF and CG intersect at point o

Because of DE//BC, so: ∠EDG=∠CBD.

Given that ∠CBD=∠CDB, then: ∠EDG=∠CDB.

And ∠ 1=∠2, and ∠ EDG+∠ CDB+∠1+∠ 2 =180.

So: ∠ 1+∠ EDG = 90.

Then in Rt△DOG, ∠BGC=50, then: ∠ dog = 40.

So: ∠ 2+∠ 4 = ∠ dog = 40 in △DOC.

And ∠ 1=∠2, ∠3=∠4, and ∠1+∠ 2+∠ 3+∠ 4+∠ dec = 65438.

So: 2 (∠ 2+∠ 4)+∠ DEC = 180.

Then: ∠ dec =180-2 (∠ 2+∠ 4) =180-80 =100.