<Tri-2025-2203>, hint

Let D' on the extended CA  such that DA=AD' 

Then BD=BD'  <BDD'=<BD'D=2<ABC. The bisector of <BD'D meets BC at M.

Since <CD'M=<ABC <D'MC=90. ( Triangles BCA and D'CM are similar)

Thus Triangles D'BM and D'CM are congruent.

Hence BD=BD'=BE+ED=BE+DC=2AD+DC. Hence BE=2AD.