Croatian Mathematical Olympiad

Since $C$ and $D$ are on $k$, the power of the point $T$ with respect to $k$ equals $|TB|^2 = |TC| \cdot |TD|$.

Furthermore, since the right-angled triangles $TBM$ and $TOB$ are similar, we have $|TB|^2 = |TM| \cdot |TO|$. Therefore, $|TC| \cdot |TD| = |TM| \cdot |TO|$, i.e. the quadrilateral $CDOM$ is cyclic.

Let point $C'$ be the intersection of $k$ and the line $DM$, while $\alpha = \angle C'MC$. Now we have
$$\angle C'C'M = \angle C'C'D = \frac{1}{2}\angle COD = \frac{1}{2}\angle CMD = \frac{1}{2}(180^\circ - \angle C'MC) = 90^\circ - \frac{1}{2}\alpha,$$
therefore $\angle MCC' = 180^\circ - \alpha - (90^\circ - \frac{1}{2}\alpha) = 90^\circ - \frac{1}{2}\alpha = \angle CC'M$, i.e. $|C'M| = |CM|$, meaning that $C'$ is the reflection of $C$ across the line $OM$, and $|AC'| = |AM|$ holds. Let $M'$ be the point on $k$ different from $C'$ such that $|AM'| = |AC'|$. Then
$$\angle M'DA = \angle C'DA = \angle MDA.$$
We can conclude that triangles $MDA$ and $M'DA$ are congruent (two pairs of congruent sides, one pair of congruent angles, and both are obtuse), so $M'$ is the reflection of $M$ across the line $AD$. The fact that $O$ is the circumcentre of the triangle $ADM'$ completes the proof.