The problem is flooded with similar triangles. Without loss of generality we will let $AC<AB$, and let $AM$ meet $KD$ at $X$. Note $OM$ and $AS$ are both perpendicular to $BC$, implying $AK||OM$. It is easy to see that $\triangle ABC$, $\triangle ASC$, $\triangle ABS$ and $\triangle ADE$ are similar. We can see from this that $\angle KAE= \angle AEK$ so that $KE=AK$ and from a symmetric argument $AK=DK$ hence $DK=KE$, i.e. $K$ is the midpoint of $DE$. Hence $\angle OKD=90^{\circ}$. Consider a circle with diameter $BC$, clearly $A$ must be on the circumference hence $MA=MB$. So $\angle MAB=\angle ABM=\angle AED$ so $\triangle ADX$ is similar to $\triangle ADE$. Hence $90^{\circ}=\angle AXD= \angle MXE $. Then both $MX$ and $OK$ are perpendicular to $DE$ i.e. $AM=MX||OK$. So $AK||OM$ and $AM||OK$.