Solution: Angle-chasing tells us $\angle ABD = \angle MDA = \angle MNA$ thus $\overarc{AQC} = \overarc{AP} + \overarc{QC}\implies \overarc{AQ} = \overarc{AP}$ so $RD$ bisects $\angle PRQ$ and $A$ is midpoint of arc $PQ$. So it suffices to show $AP=AD$ which is easy since we know $APP^2 = AB*AM = AD^2$.

similar solution but with inversion notation: 1- Angle chasing to prove that BMNC is cyclic 2- After inversion with center at A and ratio sqrt(AM*AB), P,Q and D remain fixed Done.

rmrf wrote: Solution: Angle-chasing tells us $\angle ABD = \angle MDA = \angle MNA$ thus $\overarc{AQC} = \overarc{AP} + \overarc{QC}\implies \overarc{AQ} = \overarc{AP}$ so $RD$ bisects $\angle PRQ$ and $A$ is midpoint of arc $PQ$. So it suffices to show $AP=AD$ which is easy since we know $APP^2 = AB*AM = AD^2$. How do you know that $AP^2=AB*AM$

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=516876 Sincerely Jean-Louis

Thank you so much

This problem was proposed by Burii.

∠ABC = ∠ADM = ∠ANM ---> arc AP = arc AQ ---> DR is angle bisector of PRQ. in order to prove D is incenter we need to prove AP = AD = AQ. ∠MBP = ∠MPA ---> AP^2 = AM.AB = AD^2 ---> AP = AD ∠QCN = ∠AQN ---> AQ^2 = AN.AC = AD^2 ---> AQ = AD We're Done.

Let the projections of $B$ and $C$ onto $CA, AB$ be $E, F$ respectively and $O$ be the circumcenter of $ABC$. A well-known corollary from the solution to Fagnano's Problem states that $MN$ is the perpendicular bisector of the $D$-altitude of $DEF$. Thus, we have $$PQ \equiv MN \parallel EF.$$ Because $BCEF$ is cyclic, isogonality and Reim's yields $$AO \perp EF \parallel PQ$$so $AP = AQ$. Now, observe $$\measuredangle APM = \measuredangle APQ = \measuredangle PQA = \measuredangle PBA = \measuredangle PBM$$so $$AP^2 = Pow_{(BMP)}(A) = AM \cdot AM = Pow_{(BDM)} = AD^2.$$The Incenter-Excenter Lemma finishes. $\blacksquare$ Proving the Corollary: Let $X, Y$ denote the reflections of $D$ over $M$ and $N$ respectively. Then, we can angle chase to show $E, F, X, Y$ are collinear. Other Remarks: Let $J = EF \cap BC$ and $K = MN \cap BC$. Then, considering the circle with diameter $DJ$ and center $K$ may lead to another solution, as $(DJ)$ and $(BC)$ are orthogonal. In particular, $DY$ should be related to the $R$-Mixtilinear Incircle of $PQR$.