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The excircle to side $BC$ of $\triangle ABC$ is tangent to lines $BC,AB$ and $AC$ at $D,E,F$ respectively. Let $P$ be the orthogonal projection of $D$ onto $EF$. Let the midpoint of $EF$ be $M$. Prove $ABCP$ cyclic if and only if $ABCM$ cyclic.

I didn't make much progress, except for $AD=AE$ (obviously), $AM//DP$ and $ME=MF=MI_A$ by the incircle excircle lemma. A friend told me to do root BC inversion, but that just move the excircle to the incircle/mixtilinear incircle?

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First we know that $M$ lies on the angle bisector of $\angle BAC$. Therefore if we assume $ABCM$ is cyclic, $M$ is the midpoint of arc $BC$ not containing $A$.

Let $EF$ intersect the circumcircle of $ABC$ again at $P'\neq M$. Since $AMP'=90^{\circ}$, $P'$ is the point diametrically opposite to $A$ in circle $ABC$.

Construct parallel to $AM$ at $P'$ intersecting the circle $ABC$ again at $M'$ and the line $BC$ at $D'$. $AMP'M'$ is a rectangle and thus $M'$ is the point diametrically opposite $M$ in circle $ABC$. This means $M'$ is actually the midpoint of arc $BC$ containing $A$. As a result, $\angle BP'D'=\angle CP'D'$. This combined with $D'P'$ perpendicular to $EF$, gives $K,D',C,B$ harmonic, where $K$ is the intersection of $BC$ and $EF$. ($K$ is the point that is not labelled in my diagram.) However, it is actually very well known that $(K,D;C,B)=-1$ so $D'=D$ and $P'=P$. So we are done.

The converse is the same thing: if $P$ is on circle $ABC$, $PD$ is the angle bisector of $\angle BPC$, and thus pass through the midpoint $M'$ of arc $BAC$. Line $EFP$ then intersects the circle again at $M$, the diametrically opposite point of $M'$. $M$ is then the midpoint of arc $BC$ not containing $A$. then $AM$ bisects $\angle FAE$ and hence $M$ is midpoint of $EF$.

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