Let $O$ be the circumcenter of the isosceles triangle $ABC$ ($AB = AC$). Let $P$ be a point of the segment $AO$ and $Q$ the symmetric of $P$ with respect to the midpoint of $AB$. If $OQ$ cuts $AB$ at $K$ and the circle that passes through $A, K$ and $O$ cuts $AC$ in $L$, show that $\angle ALP = \angle CLO$.
Reflect $K$ and $Q$ about $AO$, ending up at $K'$ and $Q'$. $M$ - midpoint of $AC$. $OK'=OK=OL$ $\implies ML=MK'$ $\implies PLQ'K'$ - parallelogram, $\angle ALP=\angle AK'O=\angle AKO=\angle CLO$.
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