If $K$ is the midpoint of $AH$, $P ∈ AB$, $Q ∈ AC$ and $K ∈ PQ$ such that $OK \perp PQ$, then $OP = OQ$. Proposition. Let $ABC$ be a triangle with $H$ and $O$ as it’s orthocenter and circumcenter, respectively. If $K$ is the midpoint of $AH$, $P ∈ AB$, $Q ∈ AC$ and $K ∈ PQ$ such that $OK \perp PQ$, then $OP = OQ$.

To prove this, I wish to show that $\triangle AKQ \cong \triangle HKP$, which would yield $KP = KQ$. I’ve also considered the approach of proving that $H ∈ (PKE)$, where $BH \cap AC = D$ and $OK \cap (DQK) = E$. In both cases it suffices to have $PH \parallel AC$, though I have yet to find the crux move in order to back this claim. I suspect similar triangles or the fact that $AH$ and $AO$ are isogonal to come into play, though not knowing exactly how.


Let $D$ be the intersection of $AO$ with the circumcircle. So, we have $BHCD$ is a parallelogram, and thus $BC$ bisects $HD$. Let $M$ be the intersection of $BC$ and $HD$, so $M$ is the midpoint of $HD$. Let $EF$ be the perpendicular line of $HD$ pass $H$. Since $EF\perp HD$, $BH\perp AC$, so $\angle BHC=\angle AFE$. Also, since $BD\perp AB$, $EF\perp DH$, so $\angle BDH=\angle AEF$. Thus, $\triangle AEF\sim \triangle BDH$. Also, since $M$ on $HD$ and $\angle HBM=90^\circ-\angle ACB-\angle FAH$, we also have $\triangle BHM\sim \triangle AFH$. So $HF/FE=HF/AF\times AF/FE=MH/BH\times BH/HD=HM/HD=1/2$. Therefore, $EF=2FH$, so $EH=FH$. Since $K$ is the midpoint of $AH$, $O$ is the midpoint of $AD$, we have $HD\parallel OK$. Also $OK\perp PQ$ and $HD\perp EF$, we have $PQ\parallel EF$. So $PK/KQ=EH/HF=1$. Thus, $PK=KQ$. Also, since $OK\perp PQ$, we can arrive at $OP=PQ$.
