Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

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.