Berkeley Math Circle Monthly Contest 2

Solution:

There are many ways to structure the proof. The following method seems to have minimal logical difficulties.
Because points $A$, $B$, and $C$ are not collinear, we can draw the circumscribed circle $\omega$ of $\triangle ABC$. The arc $AC$ of $\omega$, not containing $B$, is intercepted by inscribed angle $ABC$ and thus has measure $2 \angle ABC$. On this arc we may find a point $E$ such that $AE$ has the smaller measure $2 \angle ABD$. Then angles $ABD$ and $ABE$ have the same measure and orientation, so $E$ is on $BD$; also, angles $ACD$ and $ACE$ have the same measure and orientation, so $E$ is on $CD$. Since lines $BD$ and $CD$ have only one point in common, $D=E$ and thus $D$ lies on the circle.