Let $AO$ cuts $DE$ at $P,$ $Q$ is the anti-pode of $A$ wrt $(O).$ Since $DE \perp AO$, $BCED$ is cyclic Inversion at $A,$ radius $AD.AB:$ $\qquad$ $DE \leftrightarrow (O);$ $(ADE) \leftrightarrow BC$ $\Rightarrow$ $AL.AK=AD.AB=AP.AQ=AM.AO$, so $K, L, M, O$ are concyclic

electrovector wrote: In an acute $ABC$ triangle which has a circumcircle center called $O$, there is a line that perpendiculars to $AO$ line cuts $[AB]$ and $[AC]$ respectively on $D$ and $E$ points. There is a point called $K$ that is different from $AO$ and $BC$'s junction point. $AK$ line cuts the circumcircle of $ADE$ on $L$ that is different from $A$. $M$ is the symmetry point of $A$ according to $DE$ line. Prove that $K$,$L$,$M$,$O$ are circular. Clearly $BCED$ is cyclic. We have $\angle DLA = \angle DEA = \angle KBD$ or $DLKB$ is cyclic. Also $BDOM$ is cyclic (as $M$ is $A-$ antipode). Therefore we have $AK\cdot AL=AD\cdot AB = AO\cdot AM$ and we are done.