Solution:

Join \[OE\]

\[Arc\text{ }EC\] subtends \[\angle EOC\]at the centre and \[\angle EBC\]at the remaining part of the circle.

\[\angle EOC\text{ }=\text{ }2\angle EBC\text{ }=\text{ }2\text{ }x\text{ }{{65}^{o}}~=\text{ }{{130}^{o}}\]

Now, in \[\vartriangle OEC\]

\[OE\text{ }=\text{ }OC\] [Radii of the same circle]

So, \[\angle OEC\text{ }=\angle OCE\]

But, in \[\vartriangle EOC\]by angle sum property

\[\angle OEC\text{ }+\angle OCE\text{ }+\angle EOC\text{ }=\text{ }{{180}^{o}}\] [Angles of a triangle]

\[\angle OCE\text{ }+\angle OCE\text{ }+\angle EOC\text{ }=\text{ }{{180}^{o}}\]

Or,

\[2\angle OCE\text{ }+\text{ }{{130}^{o}}~=\text{ }{{180}^{o}}\]

\[2\angle OCE\text{ }=\text{ }{{180}^{o}}-\text{ }{{130}^{o}}\]

So,

\[\angle OCE\text{ }=\text{ }{{50}^{o}}/\text{ }2\text{ }=\text{ }{{25}^{o}}\]

And, \[AC\text{ }||\text{ }ED\][Given]

\[\angle DEC\text{ }=\angle OCE\] [Alternate angles]

Thus,

\[\angle DEC\text{ }=\text{ }{{25}^{o}}\]