In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D,$ such that $AD = AB$. Line segment $\overline{AD}$ is extended to $E,$ such that $CD = CE$ and $\angle DBE = \angle BAD$. Show that triangle $ACE$ is isosceles.

This is what i found “an explicit formula designates the nth term of the sequence, as an expression of n (where n = the term's location). it defines the sequence as a formula in terms of n. this example is an arithmetic sequence(the same number, 5, is added to each term to get to the next term).” i hope this you for what your looking for. can you plz mark me brainliest?