<Qua-2024-744>, hint

Let <CDB=2x <FDB=x  then <DBA=<DAB=2x

Let <ADB=y=<DAF=<AFB=<DBF

AE+EF=DE+EB=DB=DA. Triangles AFB and DBF are congruent (AF=DB, BF common,

<AFB=<DBF=y) Hence <FAB=<FDB=x...><EDA=x. <EAD=<DAB-<EAB=2x-x=x=<ADE ,y=x

Hence 5x=180 x=36  <ABD=2x=72.