We know that $$\angle AOC$$ and $$\angle BOC$$ from a linear pair
It can be written as
$$\angle BOC =180^o -80^o$$
By subtraction
$$\angle BOC =100^o$$
Using the angle sum property
$$\angle ABC +\angle BOC +\angle DCE =180^o$$
By substituting the values
$$\angle ABC +100^o +50^o +=180^o$$
On further calculation
$$\angle ABC=180^o-100^o -50^o$$
By subtraction
$$\angle ABC =180^o -150^o$$
So we get
$$\angle ABC =30^o$$