Step 1: Use angles on the same arc:
\(\angle CBE\) and \(\angle CAE\) both stand on the arc CE, so \(\angle CAE = \angle CBE = 65^\circ\).
Step 2: Use the semicircle:
AC is a diameter, so \(\angle AEC = 90^\circ\) (angle in a semicircle). In triangle AEC, \(\angle ACE = 180^\circ - 90^\circ - 65^\circ = 25^\circ\).
Step 3: Use the parallel lines:
ED \(\parallel\) AC and EC is a transversal, so the alternate angles are equal: \(\angle DEC = \angle ECA = 25^\circ\).
Final Answer:
The correct option is (1): 25.