Let $D'$ be the intersection of line $BD$ and circle $ABC$.

$ABCD'$ is cyclic. So,

\[\angle D'AC=\angle D'BC=\angle DBC=\angle ABD=\angle ABD'=\angle ACD'\]

$\therefore D'$ lies on the perpendicular bisector of $AC$ and it also lies on $BD$.

$AD=CD\implies D$ also lies on the perpendicular bisector of $AC$.

Two lines can intersect at atmost one point.

\[\therefore D=D'\]

So, $D$ is on circle $ABC$ or $ABCD$ is cyclic.