In the cyclic quadrilateral ABCD, DE is the external bisector of \(\angle BDC\). We need to prove that AE is the external bisector of \(\angle BAC\). Construction: Extend CD to point F and BA to point G. Proof: In cyclic quadrilateral AEDB, ∵ The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle ∴ \(\angle EAG = \angle EDB\) Since DE is the external bisector of \(\angle BDC\), \(\angle EDB = \angle EDF\) ∴ \(\angle EAG = \angle EDF\) — (i) Now, in cyclic quadrilateral ACDE, ∵ The exterior angle equals the opposite interior angle ∴ \(\angle EDF = \angle EAC\) — (ii) From (i) and (ii), \(\angle EAG = \angle EAC\) ∴ AE is the external bisector of \(\angle BAC\) (Proved)