Solution:
\[ABCD\]is a cyclic quadrilateral
(i) Hence, \[\angle BAD\text{ }+\angle BCD\text{ }=\text{ }{{180}^{o}}\]
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
\[\angle BCD\text{ }=\text{ }{{180}^{o}}-\text{ }{{96}^{o}}~=\text{ }{{84}^{o}}\]
And, \[\angle BCE\text{ }=\text{ }{{180}^{o}}-\text{ }{{84}^{o}}~=\text{ }{{96}^{o}}~\][Linear pair of angles]
(ii) Similarly, \[BCEF\]is a cyclic quadrilateral
So, \[\angle BCE\text{ }+\angle BFE\text{ }=\text{ }{{180}^{o}}\]
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
\[\angle BFE\text{ }=\text{ }{{180}^{o}}-\text{ }{{96}^{o~}}=\text{ }{{84}^{o}}\]