Given a triangle $ABC$, in which the medians $BE$ and $CF$ are perpendicular. Let $M$ is the intersection point of the medians of this triangle, and $L$ is its Lemoine point (the intersection point of lines symmetrical to the medians with respect to the bisectors of the corresponding angles). Prove that $ML \perp BC$.
(Mykhailo Sydorenko)
Denote $D$ by midpoint of $BC$, $E$ by intersection of tangent from $B,C$ of $(ABC)$
Since $\widehat{BMC}=90^o$ we have $DB=DC=DM=\dfrac{1}{3}DA$
$\widehat{AEB}=180^o-\widehat{ABE}-\widehat{BAE}=\widehat{C}-\widehat{DAC}=\widehat{DCM}+\widehat{ACM}-\widehat{DAC}$
$=\widehat{ACM}+\widehat{DMC}-\widehat{DAC}=2\widehat{ACM}=2\widehat{BCL}$
Combine with $EB=EC$ we have $E$ is the center of $(BLC)$
$\Rightarrow EL=EB=EC$
$\Rightarrow \dfrac{AE}{EL}=\dfrac{AE}{EB}=\dfrac{\sin \widehat{ABE}}{\sin \widehat{BAE}}=\dfrac{\sin \widehat{C}}{\sin \widehat{DAC}}=\dfrac{AD}{DC}=\dfrac{AD}{DM}$
$\Rightarrow ML \parallel DE$ or $ML \perp BC$
Thank you for posting my problem. My solution was based on isogonal conjugation in quadrilateral and angle chasing. Try showing that M and L are isogonal conjugated not only for ABC, but also for BCEF.