in an acute triangle $ABC$,$D$ is a point on $BC$,let $Q$ be the intersection of $AD$ and the median of $ABC$from $C$,$P$ is a point on $AD$,distinct from $Q$.the circumcircle of $CPD$ intersects $CQ$ at $C$ and $K$.prove that the circumcircle of $AKP$ passes through a fixed point differ from $A$.
We construct a parallelogram $XBCA$ , we are going to prove that : the circle always pass through $X$
Since $P,K,D,C$ are concyclic , we have $\angle DPK=\angle KCD = \angle KXA$ , hence $\angle APK+\angle AXK= \pi$ and $A,X,P,K$ are concyclic . we are done .
Let $X$ be such that $ACBX$ is a parallelogram. Then since $\triangle QAX \sim \triangle QDC$, $\frac{QA}{QD} = \frac{QX}{QC}$, and $CDPK$ cyclic implies $QP \cdot QD = QK \cdot QC$. Thus
\[QP \cdot QA = QP \cdot QD \cdot \frac{QA}{QD} = QK \cdot QC \cdot \frac{QX}{QC} = QK \cdot QX\]So $APBX$ cyclic.
As $D \to \infty$, $AP \parallel BC$, so the fixed point $X$ is such that $AX \parallel BC$. Next, consider $\triangle ABC$ to be equilateral, $D$ is the midpoint of $C$ and $P \to Q$, then it's obvious that $X$ lies on the $C$-median. So we can sort of guess the location of $X$.