SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) Denote $G$ as the intersection of $BE,CD$, then $G$ is the centroid of triangle $ABC$. We assume that $AB<AC$ and the solution is similar to all other cases. Since $ABLC$ is cyclic, we have $180^{\circ }-\angle AKN=\angle ACN=\angle ABM=\angle AKM$, hence $$\angle AKM+\angle AKN=180^{\circ },$$ which means $M,K,N$ are collinear. Since $AKNC$ and $AKBM$ are both cyclic, then by the power of a point to circle, we have $$LN\cdot LC=LK\cdot LA=LB\cdot LM,$$ thus $MBNC$ is concyclic and we can see $\angle CBL=\angle LNM$. Denote $P$ as the intersection of $OL$ and $MN$. Since $O$ is the circumcenter of $(ABLC)$ then $$\angle PLN=90^{\circ }-\frac{\angle LOC}{2}=90^{\circ }-\angle CBL=90^{\circ }-\angle MNL$$ Therefore, $OL$ is perpendicular to $MN$.

2) Notice that $ADGEBC$ is a complete quadrilateral and $K$ is the Miquel point so $K$ belongs to two circles ($BDG$), ($CEG$). Thus we have $$\angle DKB=\angle DGB=\angle EGC=\angle EKC,\angle KDB=\angle KGB=\angle KCE.$$ So $△BKD\sim △EKC$, then by sin law, we have $$\frac{AB}{AC}=\frac{BD}{CE}=\frac{BK}{KE}=\frac{\sin BAK}{\sin EAK}=\frac{\sin BAL}{\sin CAL}$$ which means $ABLC$ is the harmonic quadrilateral and $LA$ is the symmedian of triangle $LBC$. Finally, because $MN$ is the antiparallel to $BC$ with respect to $\angle BAC$ then the symmedian of triangle $LBC$ is the median of triangle $LMN$, which means $LA$ passes through the midpoint of $MN$ or $K$ is the midpoint of the segment $MN$. $\square$