Balkan Mathematical Olympiad

Let $K$ be the second intersection point of $AD$ with ${\Gamma }_1$, $Q$ the second intersection of $BC$ with ${\Gamma }_2$, $P$ the intersection of $AC$ and ${\Gamma }_1$, and $G$ the intersection of $BD$ and ${\Gamma }_2$. We will show that $K$, $G$, $P$, $Q$ all lie on a line perpendicular to both $AD$ and $BC$.

Since $MNBQ$ and $KMNA$ are cyclic quadrilaterals, we have $$DB\cdot DG=DM\cdot DN=DA\cdot DK.$$ By the power of a point theorem, $ABGK$ is cyclic and $\angle AKG=\pi -\angle ABG=\frac{\pi }{2}$. But $AP\perp l=AB\Rightarrow \angle AKP=\angle PAB=\frac{\pi }{2}=\angle AKG$, hence $K$, $P$ and $G$ are collinear. Similarly, we show that $G$, $K$, $Q$ are collinear. Indeed, since $PMNA$ and $QMNB$ are cyclic quadrilaterals, we have $$CP\cdot CA=CM\cdot CN=CQ\cdot CB.$$ Therefore, $PQBA$ is cyclic and $\angle PQB=\pi -\angle PAB=\frac{\pi }{2}$. Since also $\angle GQB=\frac{\pi }{2}$, it follows that $G$, $P$, $Q$ are collinear. Hence $K$, $P$, $G$, $Q$ are collinear and $KQ\perp AD$, $KQ\perp BC\Rightarrow AD\parallel BC$. Since $AC\parallel BD\perp AB$, $ACBD$ is a parallelogram.