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