Solution:
Join \[EB\]
Then, in cyclic quad. \[ABEP\]
\[\angle APE\text{ }+\angle ABE\text{ }=\text{ }{{180}^{o}}~\ldots ..\text{ }\left( i \right)\][Opposite angles of a cyclic quad. are supplementary]
Similarly, in cyclic quad.BCQE
\[\angle CQE\text{ }+\angle CBE\text{ }=\text{ }{{180}^{o}}~\ldots ..\text{ }\left( ii \right)\][Opposite angles of a cyclic quad. are supplementary]
Adding (i) and (ii), we have
\[\angle APE\text{ }+\angle ABE\text{ }+\angle CQE\text{ }+\angle CBE\]
\[~=\text{ }{{180}^{o~}}+\text{ }{{180}^{o}}~=\text{ }{{360}^{o}}\]
Or,
\[\angle APE\text{ }+\angle ABE\text{ }+\angle CQE\text{ }+\angle CBE\text{ }=\text{ }{{360}^{o}}\]
But, \[\angle ABE\text{ }+\angle CBE\text{ }=\text{ }{{180}^{o}}~\][Linear pair]
\[\angle APE\text{ }+\angle CQE\text{ }+\text{ }{{180}^{o}}~=\text{ }{{360}^{o}}\]
\[\angle APE\text{ }+\angle CQE\text{ }=\text{ }{{180}^{o}}\]
Therefore, \[\angle APE\text{ }and\angle CQE\]are supplementary.