Pretty much one Pascal and you don't need a huge part of the assumptions.

Assassino9931 wrote: Pretty much one Pascal and you don't need a huge part of the assumptions. BMO moment

We can generalize the problem as follow: Quadrilateral $ABCD$ . Diagonals $AC$ cut $BD$ at $P$ and suppose $AB$ cut $CD$ at $E$. $X,Y$ lie on $EB,EC$ such that $EXPY$ is a parallelogram. $M$ is the midpoint of diagonals $EP,XY$. $BM,CM$ cut $(EBC)$ at $P,Q$ .Prove that $(PMY),(QMX)$ intersects on $(EBC)$.

Very nice problem Let $M$ be the midpoint of $BC$ and let $MH$ intersect $(ABC)$ at $T$. We claim that both circles pass through $T$. The pith of the proof is the following Claim. Claim: Quadrilateral $ATPQ$ is cyclic. Proof: It suffices to prove that $\angle PTQ=\angle A$, or equivalently that $\angle PTQ=\angle BTC,$ that is $\angle BTP=\angle QTC$. Therefore, it suffices to prove that triangles $BTP$ and $CTQ$ are similar. Since $\angle TBP=\angle TCQ$, we are left to prove that $\dfrac{BP}{CQ}=\dfrac{TB}{TC}$. Let $E,F$ be the midpoints of $AB,AC$ respectively. Then, $KE \parallel BH$ and $OF \perp AC,$ and so $KE \parallel OF$. Similarly, $KF \parallel OE,$ and so $KEOF$ is a parallelogram. Therefore, $\angle KPO=\angle KEO=\angle KFO=\angle KQO,$ hence triangle $OPQ$ is isosceles, i.e. $K$ is the midpoint of $PQ$. Thus, quadrilateral $APHQ$ is a parallelogram, sicne $K$ is the midpoint of both $PQ$ and $AH$. Therefore, $\angle BPH=\angle BAC=\angle HQC,$ and $\angle PBH=\angle QCH$. Hence, triangles $BPH$ and $CQH$ are similar. Thus, if we let $A'$ be the antipode of $A$ in $(ABC)$, $\dfrac{BP}{CQ}=\dfrac{BH}{CH}=\dfrac{CA'}{BA'}=\dfrac{BT}{TC},$ as desired (the middle equality follows since $BHCA'$ is a parallelogram, and the last one from an easy application of the Ratio Lemma) $\blacksquare$ To the problem, note that from the Claim we have $\angle TPK=180^\circ-\angle TAC=180^\circ-\angle TYK,$ and so $T \in (KPY)$. Similarly, $T \in (KQX)$, and so we conclude.

Quite a silly problem (as #2 also mentions) in the sense that the following stronger problem holds: Let $K$ be any point inside $\triangle ABC$ and $\ell$ be any line through $K$, intersecting sides $AB$ and $AC$ at $P, Q$. If $X, Y$ are defined as the second intersections of lines $BK$ and $CK$ with $\omega$ (the circumcircle of $\triangle ABC$), then the circumcircles of $\triangle KPY$ and $\triangle KQX$ intersect on $\omega$. Here's a solution to that problem: If $Z$ is the second intersection of $\overline{XQ}$ and $\omega$, then Pascal's theorem on $BXZYCA$ implies that $Z, P, Y$ are collinear. If $T$ is the second intersection of the circumcircle of $\triangle KQX$ and $\omega$, then (using directed angles): \[\angle PYT = \angle ZYT = \angle ZXT = \angle QXT = \angle PKT \]Hence $PKTY$ is cyclic and we're done.

This problem was proposed by Stefan Lozanovski.

If $M$ is the midpoint of $BC$ then $KO \parallel HM$, and since $AK \perp BC, AQ \perp BH, AP \perp CH$, $AK$ bisects $PQ$. Thus, $\angle{BHP} = \angle{CHQ} = 90^{\circ} \implies$ if $J = (APQ) \cap (ABC)$, then a spiral similarity at $J$ takes $PQ$ to $BC$, and if $AH \cap (ABC) = H_A$ then $BJCH_A$ is harmonic, so $\triangle{JPK} \sim \triangle{JBM} \sim \triangle{JH_AC} \implies J$ is the center of the spiral similarity taking $PK$ to $H_AC$. Thus, $(JPK) \cap (JH_AC) = PH_A \cap CK = Y \implies JYKP$ is cyclic, and similarly $JXQK$ is cyclic, so we are done. $\square$