XXXIII wrote: Let $L$ be the point on $BC$ such that \( AL \) is the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). Proposed by Mykhailo Plotnikov the problem statement isn't complete, $L$ should be lying on $BC$ so that $PQ\perp AO$. anyway, complex bash seems promising to do.

Let $T$ be the point on $BC$ such that $TA$ is tangent to $(ABC)$. Note that $TB\times TC=TA^2=TL^2$, so $\frac{TB}{TL}=\frac{TL}{TC}$. This means $\frac{TB}{BL}=\frac{TL}{LC}$ so $\frac{TB}{BD}=\frac{TL}{LE}$. Take $L'$ on segment $TA$ such that $\frac{TL'}{L'A}=\frac{TB}{BD}=\frac{TL}{LE}$, then $L'B\parallel AD$ and $L'L\parallel AE$. Since $L'B\parallel PD$ and $BD=DL$, hence $LP=L'P$. Let $M$ be the midpoint of $AL$. There is a homothety (with $r=0.5$) from $L$ taking $L'ACB$ to $PMED$. Hence, $PM\parallel L'A$. But $L'A\perp AO$, so we are done.