Let $QX$ and $PY$ be tangent to the circumcircle at $D$ and $E$. Since $\angle BCM =\angle BAN=\angle BMQ$, we have that $QM$ is tangent to $(BCM)$. Then $QD^2=QB \cdot QC=QM^2$, so $QD=QM$ and similarly $PE=PM$. Finally, $QZ-PZ=QZ+ZD-PZ-ZE=QD-PE=QM-PM$, so the quadrilateral $QMPZ$ is circumscribed, and so $MXZY$ is also circumscribed.

Mindstormer wrote: Let $QX$ and $PY$ be tangent to the circumcircle at $D$ and $E$. Since $\angle BCM =\angle BAN=\angle BMQ$, we have that $QM$ is tangent to $(BCM)$. Then $QD^2=QB \cdot QC=QM^2$, so $QD=QM$ and similarly $PE=PM$. Finally, $QZ-PZ=QZ+ZD-PZ-ZE=QD-PE=QM-PM$, so the quadrilateral $QMPZ$ is circumscribed, and so $MXZY$ is also circumscribed. I think your solution is wrong!!

Mathuzb wrote: Mindstormer wrote: Let $QX$ and $PY$ be tangent to the circumcircle at $D$ and $E$. Since $\angle BCM =\angle BAN=\angle BMQ$, we have that $QM$ is tangent to $(BCM)$. Then $QD^2=QB \cdot QC=QM^2$, so $QD=QM$ and similarly $PE=PM$. Finally, $QZ-PZ=QZ+ZD-PZ-ZE=QD-PE=QM-PM$, so the quadrilateral $QMPZ$ is circumscribed, and so $MXZY$ is also circumscribed. I think your solution is wrong!! I think his solution is right

Actually Mindstormer's solution is right.

After getting $PM=PE$ and $QM=QD$, there's another way to finish the problem (lengths are hard) Let $X$ be the intersection of the angle bisectors of $\angle MPE$ and $\angle MQD$. It suffices to show that $MX$ bisects $\angle PQM$, since then your quadrilateral would be circumscribed. To do this, notice that $X$ is actually just the circumcenter of $(MDE)$, and at this point there's tonnes of circles passing through $M$, so let's invert about $M$ to get the following problem: Problem: $ABC$ triangle, $P,Q$ such that $MACP$ and $MBCQ$ are both isosceles trapeziums in that order, $D,E$ the intersections of the perpendicular bisectors of $MQ,MP$ with the circumcircle (so that $D,E$ are on the opposite sides as $C$ in lines $MQ$ and $MP$ I presume). Show that $DE$ is perpendicular to the angle bisector of $\angle PMQ$ (since the circumcenter inverts to some point on the line through $M$ perpendicular to $DE$). The problem is kinda trivial now, since the isos traps show that $D,E$ actually just lie on the perpendicular bisectors of $BC$ and $AC$ respectively, so $DE$ is perpendicular to the internal angle bisector of $\angle C$, and now we just have two angle bisectors of a parallelogram.

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Is the "MEDIAN" condition redundant?

Yes, the median condition is not necessary, which makes the question more difficult imo

Actually that M on BC suffices.

Mindstormer wrote: the quadrilateral $QMPZ$ is circumscribed, and so $MXZY$ is also circumscribed. I realize it's quite obvious but could you please prove that or give the idea of a proof to me?

Why $CM$ median ???

$PM = QL$ $QM = PK$ It suffices to prove that $PM + QZ = QM + PZ$

Solved with Shreyasharma. Rather instructive problem. First time I ever saw this Pitot's thing. We first make the following observations. Claim : Line $QM$ is tangent to $(BCM)$ and $PM$ is tangent to $(ACM)$. Proof : Simply note that, \[\measuredangle BMQ = \measuredangle BAN = \measuredangle BCN = \measuredangle BCM \]Thus, $QM$ is indeed tangent to $(BCM)$. Similarly, we obtain that $PM$ is also tangent to $(ACM)$. Now, by Excentral Pitot's Theorem, it suffices to show, \[MP + QZ = MQ + PZ\]Let $D,E$ be the tangency points of $QX$ and $PY$ with $(ABC)$. We now note that, \[QZ-PZ = (QD-DZ) - (PE-EZ)= QD-PE\]since $ZD=ZE$ by the Two Tangents Theorem. Further, \[QD^2 = QA \cdot QB = QM^2\]This allows us to conclude that $QD=QM$. Similarly, we obtain that $PE=PM$. Thus, \[QZ-PZ= QD-PE = QM-PM\]from which we conclude that \[MP+QZ=MQ+PZ\]and indeed $MXYZ$ is circumscribed, as desired.