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Suppose $\omega$ is a circle with center $O$ and $DE$ is a chord of it. The tangents of $\omega$ at $D$ and $E$ intersect each other at $P$. $M$ is the midpoint of $DE$. A line through $M$ intersects $\omega$ at $X$ and $Y$. Prove that the incenter of $\triangle PXY$ lies on $\omega$.
Hint:

Me and Zadid did this one in Zadid's home. I used total symmetry and Zadid used the killer lemma (you know which one).
Zadid's solution was more beautiful than mine though. :-/

On a second thought, as this one's your self made one, me and Zadid may have solved very similar one to this one. But the ideas are same. Sorry for the inconvenience. :-/

Tahmid Hasan wrote:তোর সমস্যাটা কী?

সব ভুইলা যাই।

In $ODPE$ , $\angle ODP=\angle OEP=90^o$ , $ODPE$ is cyclic quadrilateral.
$OM.MP=DM.ME=XM.MY$ , therefore $O,Y,P,X$ are concyclic.
$OX=OY \Rightarrow \angle OXY=\angle OYX\Rightarrow \angle OPY=\angle OXY$ ; let $OP$ intersects $\omega $ at $J$. $JP$ bisects $\angle XPY$.
on the other hand , $\angle XYP=\angle XOP=2\angle XYJ$. $JY$ bisects $\angle XYP$.
Then it is proved $J$ is the incenter of $\Delta PXY$.