I had heard that $K$ could be replaced with any point on the perpendicular bisector of $BC$.

I can't geo apparently wow why didn't I just do miquel stuff kill me pls

So there WAS a angle chasing solution... Sketch of my complicated solution : 1. Let the Euler reflection point($X_{110}$) of triangle $ABC$ be $X$. Then $Q ,X$ are symmetric WRT line $OK$. 2. Let the Euler line of triangle $ABC$ cuts $BC$ at $J$, and let $J' \in BC$ such that $\overrightarrow{CJ}=-\overrightarrow{BJ'}$. If $T=AK \cap BC$, then $T,K,P,Q$ are concyclic. 3. Chase angles!

Since $(C,E,Q,P)$ are concyclic, Chasing angles kills the problem. (Let $L=PQ \cap$ (the circle with diameter $AK$))

Let $(PQ$ intersect the circle with diameter $(AK)$ in $X$. We have $\angle QED=\angle QAD=\angle QCB$, so $P,Q,E,C$ are concyclic. But $\angle ECP=C$, so $\angle XQE=\angle SXCE=C$. So it's enough to show that $\angle OKE=\angle C$ if we want $X-O-K$ to be collinear. Let $M$ be the midpoint of $BC$. $\angle KMC=\angle KEC=90$ so $K,M,E,C$ are concyclic. This implies $\angle MCE=\angle MKE=C$. The conclusion follows.

Let $PQ$ intersect $(ADE) $ at $N $.Since $Q$ is Miquel point of $BDEC $ we have $CQEP$ is cyclic.Then easy angle chasing yields that $AN\parallel BC$ so done.

A nice extension : If we let $T$ be the isogonal conjugate of $K$ wrt $\triangle{ABC}$, then the orthocenter of $\triangle{TBC}$ lies on $KQ$.

Wow Iminsl! You solved it so hard.

Is $K$ actually important? lminsl wrote: Triangle $ABC$ satisfies $\angle ABC < \angle BCA < \angle CAB < 90^{\circ}$. $O$ is the circumcenter of triangle $ABC$, and $K$ is the reflection of $O$ in $BC$. $D,E$ is the foot of perpendicular line from $K$ to line $AB$, $AC$, respectively. Line $DE$ meets $BC$ at $P$, and a circle with diameter $AK$ meets the circumcircle of triangle $ABC$ at $Q(\neq A)$. If $PQ$ cuts the perpendicular bisector of $BC$ at $S$, then prove that $S$ lies on the circle with diameter $AK$. Note that $Q$ is the spiral center $\overline{DE} \mapsto \overline{BC}$. Now $\angle DQP=\angle DBC$ since $BDQP$ is cyclic. Along with $ADKES$ cyclic we get $\angle DQS=\angle BAS=\angle A+\angle C$ so $\angle DQP+\angle DQS=180^{\circ}$. Thus, $\overline{PQ}$ passes through $S$. Done.

lminsl wrote: I had heard that $K$ could be replaced with any point on the perpendicular bisector of $BC$. lminsil, i think thay k can be replaced with any point in the triangle

We prove a much more general statement as follows:- Generalisation wrote: Let $D$ and $E$ be points on segments $\overline{AB}$ and $\overline{AC}$ in an acute-angled triangle $\triangle ABC$. Let $\overline{DE} \cap \overline{BC}=P$ and $\odot (ADE) \cap \odot (ABC)=Q$. Suppose $\overline{PQ}$ meets $\odot (ADE)$ at $X$. Then show that $\overline{AX} \parallel \overline{BC}$. Invert about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the angle bisector of $\angle BAC$. Then we are left with the following equivalent problem- Inverted Problem wrote: Let $D$ and $E$ be points on segments $\overline{AB}$ and $\overline{AC}$ in an acute-angled triangle $\triangle ABC$. Let $\overline{DE} \cap \overline{BC}=P$ and $\odot (ADE) \cap \odot (ABC)=Q$. Suppose $\overline{DE}$ meets $\odot (APQ)$ at $X$. Then show that $\overline{AX}$ is tangent to $\odot (ABC)$. Note that, as $Q$ is the center of spiral similarity that takes $\overline{DE}$ to $\overline{BC}$, so we get that $$\measuredangle QEP=\measuredangle QED=\measuredangle QCB=\measuredangle QCP \Rightarrow CEQP \text{ is cyclic.}$$Using the above, we have $$\measuredangle XAQ=\measuredangle XPQ=\measuredangle EPQ=\measuredangle ECQ=\measuredangle ACQ \Rightarrow AX \text{ is tangent to } \odot (ABC) \text{. } \blacksquare$$

Let $SE\cap BC=X.$ After a simple angle chasing one can easily prove that $BDEX$ and $QSBX$ are cyclic quadrilaterals. Knowing this, the desired result follows after applying the radical axes theorem on $(BDEX), $ $(QSBX),$ and $(QSDE).$

Let $M$ be the midpoint of $BC.$ Since $Q$ is the Miquel point of $CEDB, CEQP$ is cyclic, and we also know that $ECKM$ is cyclic because $\measuredangle CEK = \measuredangle CMK = 90^\circ.$ As a result, we then get that $\measuredangle EKS = \measuredangle ECP = \measuredangle EQS,$ so $KEQS$ is cyclic as desired. $\blacksquare$

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Clearly, $Q$ is th center of spiral similarity that sends, $DE$ to $BC$. Therefore, $PQEC$ is cyclic. Let, $S' = OK\cap (AQCKD)$. Clearly, $AS'||BC$. Now, $ \measuredangle EQP =\measuredangle ECP =\measuredangle EAS' =\measuredangle EQS' $ So, $P,Q,S'$ are colinear and this implies $S'\equiv S$ and we are done. $\blacksquare$

$Q$ is the Miquel Point of $(BCED)$. Then $$\angle QSK = 90^\circ - \angle QPC = 90^\circ - \angle QEA = 90^\circ - \angle QKA = \angle KAQ,$$so $S$ lies on the circle with diameter $\overline{AK}$.

rmtf1111 wrote: A nice extension : If we let $T$ be the isogonal conjugate of $K$ wrt $\triangle{ABC}$, then the orthocenter of $\triangle{TBC}$ lies on $KQ$. How to prove this claim?