Saudi Arabia Mathematical Competitions

Let $P$ be the second point of intersection of segment $BC$ and the circle circumscribed about quadrilateral $AKLM$. Denote by $E$ the intersection point of the lines $KN$ and $BC$ and by $F$ the intersection point of the lines $MN$ and $BC$.

Then $\widehat{BCN}=\widehat{BAN}$ and $\widehat{MAL}=\widehat{MPL}$, as angles on the same arc. Since $AL$ is a bisector, $\widehat{BCN}=\widehat{BAL}=\widehat{MAL}=\widehat{MPL}$, and consequently $PM\parallel NC$. Similarly we prove $KP\parallel BN$. Then the quadrilaterals $BKPN$ and $NPMC$ are trapezoids; hence $$S_{BKE}=S_{NPE}\text{and}{S}_{NPF}=S_{CMF}.$$ Therefore $S_{ABC}=S_{AKNM}$.