Let $\angle ABC=\beta$. Then, $$ \angle AOC=2\beta \Longrightarrow \left\{\begin{array}{lll} \angle CAO=90^o-\beta \Longrightarrow \angle QAC=\beta \\ \angle OCA=90^o-\beta \Longrightarrow \angle ACQ=\beta \end{array} \right. $$Therefore, lines $AQ$ and $BQ$ are tangent to $\Omega$ at $A$ and $C$, respectively. Now, let $P=MB\cap QC$. Since $MP\parallel AC$, then $$ \angle MPC=\angle ACQ=\beta \, ; \, \angle AMP=\angle QAC=\beta $$Since $AMBN$ is a parallelogram, then $$ \angle CNB=180^o-\angle AMB=180^o-\beta $$As a consequence, $$ \angle CNB+\angle BPC=180^o, $$and so $$ \text{quadrilateral} \quad BPCN \quad \text{is cyclic.} \quad (1) $$Since line $AM$ is tangent to $\Omega$, then $$ \angle BAM=\angle BCA=\gamma $$ Now, let's consider the points $A,B,C$ as points belonging to the sides of the triangle $\triangle QMP$. Let's consider the Miquel point of such configuration, and let's name it $K$. I claim that points $M,K,N$ are collinear. $$ K\in (BPC)\stackrel{(1)}=(BPCN) \Longrightarrow \angle NKB=180^o-\angle BCN=180^o-\gamma \quad (2) $$Also, $$ K\in(AMB) \Longrightarrow \angle BKM=\angle BAM=\gamma \quad (3) $$From $(2)$ and $(3)$ we get that $M,K,N$ are collinear. Finally, let's prove that $B,K,Q$ are collinear. $$ \left\{\begin{array}{lll} K\in (AMB) \Longrightarrow \angle BKA=180^o-\angle AMB=180^o-\beta \\ K\in (CQA) \Longrightarrow \angle AKQ=\angle ACQ=\beta \end{array} \right. \Longrightarrow \angle BKA+\angle AKQ=180^o $$Therefore, $B,K,Q$ are collinear. The problem is solved, because we have proved that, one same point, $K$, satisfies that $K\in (AOC)$, points $M,K,N$ are collinear; and points $B,K,Q$ are collinear.

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