Question 979406
This has to be drawn.
ABC is a 1,1,sqrt(2) right triangle.  Draw it with C, the right angle, at the vertex, and CA and CB heading down at 45 degree angles to the left and to the right.  CD is parallel to AB and is drawn to the right.  

If I make AB sqrt (2), the hypotenuse, CD is sqrt (2). 
CB=AB= 1.  The distance between the lines uses another 45-45-90 right triangle, where AC is the hypotenuse.
AC is 1, so the distance between the parallel lines is sqrt(2)/2.



Now, using point E, which makes a right angle with BE and DE, we have a third triangle, with hypotenuse sqrt(2) and DE sqrt(2)/2.  This makes it a 30-60-90 right triangle, where the short side is half the hypotenuse in a right triangle.
That makes DBE at 30 degree angle.  CBA is a 45 degree angle, and DBC must be the supplement or 105 degrees.