I think there is a typo? MazeaLarius wrote: Let $D,E$ and $T$ be the intersections of $\omega$ and $AB,AC$ and $\omega$ Probably $T$ is the $A$-mixtilinear touchpoint, but I'm not sure where $P$ lies on: $\omega$ or $\Omega$?

I guess $P$ is on $\omega$ only, while $T$ is the $A$-mixtilinear intouch point. Here's the solution: Let $X,Y=\omega \cap BC$. It's well known that $TX,TY$ are isogonal in $\angle BTC$ (for a proof, try using Steiner's Ratio Theorem along with power of points $B$ and $C$ wrt $\omega$), and that $TI$ passes through the arc midpoint of $\widehat{BAC}$ in $\Omega$. In particular, $TI$ bisects $\angle XTY$, and so $P$ is the midpoint of $\widehat{XY}$ in $\omega$. Then Shooting Lemma gives $PD \cdot PM=PE \cdot PN$, as desired.

anantmudgal09 wrote: I think there is a typo? MazeaLarius wrote: Let $D,E$ and $T$ be the intersections of $\omega$ and $AB,AC$ and $\omega$ Probably $T$ is the $A$-mixtilinear touchpoint, but I'm not sure where $P$ lies on: $\omega$ or $\Omega$? Yes, sorry. I modified.

Thanks! I'll post my sketch here since I believe one can entirely angle chase it? We need to show $\measuredangle(BC, PD)=\measuredangle (PE, ED)$. By homothety at $T$, we see that $D, E$ map to arc midpoints $M_C, M_B$ opposite $C$ and $B$; and $P$ to $M_a$ the arc midpoint of $\widehat{BAC}$. Chasing angles, it suffices to show that $\measuredangle M_BM_CM_a-\measuredangle M_aM_BM_C=\measuredangle(BC, M_BM_C)$. Both sides equal $\tfrac{1}{2}(\angle B-\angle C)$ in measure, and we're done.

Here's my solution : Consider $\mathcal{H}$ the homothety centered at $T$ which carries $\omega$ to $\Omega$, and let $\{K\}=TB \cap \omega $, $\{L\}=TC \cap \omega$, and $\{T'\}=TI \cap \Omega$. We have that $\mathcal{H}(K)=B$, $\mathcal{H}(L)=C$, $\mathcal{H}(P)=T'$, but since $T'$ is the midpoint of $\overarc{BAC}$, it results that $P$ is the midpoint of $\overarc{KPL}$ $(1)$. Also, we can see that $\mathcal{H}$ sends $KL$ to $BC$, thus $KL \parallel BC$ $(2)$. Now, let $\{R,S\}=BC \cap \omega$. Using $(1)$ and $(2)$ it follows that $P$ is the midpoint of $\overarc{RPS}$. To finish, we will prove that $\angle PMN= \angle PED$ : $\angle PMN=\frac{1}{2} (\overarc{PS}-\overarc{RD})=\frac{1}{2}(\overarc{PR}-\overarc{RD})=\frac{1}{2}\overarc{PD}=\angle PED$, and the conclusion follows. $\square$ Remark: The circumcenter of the quadrilateral $DENM$ is the intersection of $AI$ with the perpendicular bisector of $[MN]$.

The center is $A$-excircle center.

Does anyone have a solution to prove the second part?

Prove that $BD=BM,\ CE=CN$.

Let $P'$ be the topmost point of $\Omega$ and let $D' = CI \cap \Omega , E' = BI \cap \Omega$. Let $M' = P'D' \cap BC , N' = P'E' \cap BC$. By a homothety at $T$, it suffices to show that $M' , D' , E' , N'$ are concyclic. Note that $P'D' \parallel BI$ since $\angle P'D'I = 90 - \angle A/2 = \angle BID'$. Similarly, $P'E' \parallel CI$. And so $\angle D'M'N' = \angle B/2 = \angle P'E'D'$ , implying that $M' , D' , E' , N'$ are concyclic. Now finally note that $\angle BDM = \angle ADP = \angle DTP = \angle PED = \angle DMB$ so $BM = BD$ and similarly $CN = CE$. So the center of $(DEMN)$ lies on both exterior bisectors of $\angle B , \angle C$, thus it is the excenter.

After proving that $D, E, N, M$ lie on a circle, I will prove $I_A$ (the $A$-excentre) is the centre of $(DEMN)$. Let $X$ be the projection of $I_A$ to $BC$ and $Y$ the projection of $I$ to $BC$. We have that $I_A$ is on the perpendicular bisector of $DE$ (line $AI$). I will prove that $I_A$ is on the perpendicular bisector of $MN \iff X$ is the midpoint of $MN$ $MX=XN \iff BM+BX=NC+CX \iff DB+BX = CE+CX \iff BD-CE = CX-BX = CX-CY = -XY$ So, I want $XY = CE-DB = (CE+AE) - (DB+AD)=AC-AB=b-c$ ($p=\frac{a+b+c}{2}$) $XY=BX-BY = (p-c) - (p-b) = b-c \implies XY=BD-CE \implies X$ is the midpoint of $MN$ So, $I_A$ is on the perpendicular bisectors of $MN$ and $DE$ and because $D, E, M, N$ lie on the same circle, $I_A$ is the centre of this circle $\blacksquare$