Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle.

This was quite classical. Let $M$ be the midpoint of major arc $BC$. We split the proof into two Claims. Claim 1: Lines $MI$ and $DN$ meet on $\omega$. Proof: Let $DN$ intersect $\omega$ at point $S$. Since $\angle DIB=\angle AIB-\angle AIE=(90^\circ+\dfrac{\angle C}{2})-90^\circ=\dfrac{\angle C}{2}=\angle ICB,$ we obtain that $DI$ is tangent to circle $(BIC)$. Thus, $DI^2=DB \cdot DC=DS \cdot DN,$ and so $IS$ is perpendicular to $DN$, that is $\angle ISN=90^\circ=\angle MSN$, hence points $M,I,S$ are collinear, as desired $\blacksquare$ Claim 2: Points $M,I,Q$ are collinear. Proof: Let $MI$ intersect $BC$ at point $Q'$. Then, by spiral similarity and the fact that $PQ$ bisects $\angle BPC$, we obtain $\dfrac{BQ}{QC}=\dfrac{BP}{PC}=\dfrac{BE}{CF}$ Now, by Menelaus' theorem and the fact that $DI$ is tangent to $(BIC)$, we easily obtain that $\dfrac{BE}{CF}=\dfrac{DB}{DC}=\dfrac{BI^2}{CI^2}$ Now, note that by an application of the ratio lemma and some sine laws, $\dfrac{BQ'}{Q'C}=\dfrac{BM}{MC} \cdot \dfrac{\sin \angle BMI}{\sin \angle CMI}=\dfrac{BI^2}{CI^2},$ hence the two ratios $\dfrac{BQ}{QC}$ and $\dfrac{BQ'}{Q'C}$ are equal, and so $Q \equiv Q'$, as desired $\blacksquare$ To the problem, combining both claims we are done.

Similar to https://artofproblemsolving.com/community/c6h476508p2667975, Iran TST 2012 Day 1Q2

PNT wrote: Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle. First let's prove that $DN\cap \omega =T$ where $T$ is the point of tangency of the mixtilinear incircle. Let $DN\cap \omega =T'$ and $M$ the midpoint of arc $BAC$ then we want to prove that $M,I$ and $T'$ are collinear or equivalently $\angle BT'I=90^{\circ}$, but $\angle DIN=90^{\circ}$ so it suffices to prove $DT'\cdot DN=DI^2$. We have $$DT'\cdot DN=DB\cdot DC=DI^2$$That's because $DI$ is tangent to $(BIC)$. Therefore $T'=T$. Note that $\angle DPN=\angle DPT+\angle TPN=\angle MNT+\angle TMN=90^{\circ}$ (We proved this above) that means $D,P$ and $M$ are collinear. Now define $Q'=MT\cap BC$, proving $Q=Q'$ will finish the problem. Since $$-1=(B,C;N,M)\overset{P}{=}(D,Q;B,C)\overset{M}{=}(P,T'';B,C)$$That means $PT''$ and $PN$ meet on $BC$ so $Q=Q'$.

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PNT wrote: PNT wrote: Hint : Prove that the intersection point is the point of tangency of the mixtilinear incircle. First let's prove that $DN\cap \omega =T$ where $T$ is the point of tangency of the mixtilinear incircle. Let $DN\cap \omega =T'$ and $M$ the midpoint of arc $BAC$ then we want to prove that $M,I$ and $T'$ are collinear or equivalently $\angle BT'I=90^{\circ}$, but $\angle DIN=90^{\circ}$ so it suffices to prove $DT'\cdot DN=DI^2$. We have $$DT'\cdot DN=DB\cdot DC=DI^2$$That's because $DI$ is tangent to $(BIC)$. Therefore $T'=T$. Note that $\angle DPN=\angle DPT+\angle TPN=\angle MNT+\angle TMN=90^{\circ}$ (We proved this above) that means $D,P$ and $M$ are collinear. Now define $Q'=MT\cap BC$, proving $Q=Q'$ will finish the problem. Since $$-1=(B,C;N,M)\overset{P}{=}(D,Q;B,C)\overset{M}{=}(P,T'';B,C)$$That means $PT''$ and $PN$ meet on $BC$ so $Q=Q'$. You assumed the collinearity when you stated that $\angle DPT = \angle MNT$, here's a proof for the collinearity: Notice that $P$ is the center of the homothety sending $BE$ to $CF$, and since $D$ is the intersection of $(EF)$ and $(BC)$ then $PEBD$ IS cyclic, thus $\angle PDI= \angle PDE = \angle PBE = \angle PBA = \angle PNA = \angle PNI$ thus, $PIND$ is cyclic and this gives us $\angle DPN = 90$.

$\angle DIB=\angle \frac{C}{2}=\angle ICB \implies DI^2=DB.DC=DK.DN \implies \angle IKD=90$ Let's prove that $QK \perp DN$ $\angle APE=180-\frac{B+C}{2}$ and $\angle APB=180-C \implies \angle EPB=\frac{B-C}{2}=\angle EDB$ $\implies P,E,D,B$ are cyclic. $\angle PNI=\angle PNA=\angle PBA= \angle PBE=\angle PDE=\angle PDI \implies P,D,N,I$ are cyclic. So $\angle DPN=90$ Also $\angle BPN=\angle \frac{A}{2}=\angle BPN,$ $NK.ND=NI^2=NB^2=NQ.NP$ which means $P,D,K,Q$ are cyclic. $\implies QK \perp DN$

We redefines everything Let $T$ be mixtilinear Incircles touch point.Note that circle pass though $E$ and $F$ Define $Y$ be antipode of $N$ then we know $\overline{Y-I-T}$ and $\angle YTN = 90$ Let $D = NT \cap BC$ and $M$ be midpoint of $BC$. Redefine $Q = DM \cap YT$ Now note $Q$ is orthocenter of $\triangle YND$. Redefine $P = DY \cap (ABC)$ now we get $NP \perp DY$ Now if we prove $P$ lie on $(AEF)$ we are done [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.603703365062785cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.889434907857462, xmax = 42.71426845720532, ymin = -19.217449558241984, ymax = 11.666426980388458; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw((4.676888221431007,6.213756330118006)--(1.46,-5.72)--(19.64,-5.64)--cycle, linewidth(0.8) + zzttqq); draw((10.489221436718749,8.131928505665023)--(-6.243576487131332,-5.753899126456025)--(10.576325071104362,-11.662372408466327)--cycle, linewidth(0.4) + qqwuqq); 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draw((-6.243576487131332,-5.753899126456025)--(19.64,-5.64), linewidth(0.8)); /* dots and labels */ dot((4.676888221431007,6.213756330118006),dotstyle); label("$A$", (4.2985779297551305,6.610825809810868), NE * labelscalefactor); dot((1.46,-5.72),dotstyle); label("$B$", (0.7906097705788365,-6.561250709449108), NE * labelscalefactor); dot((19.64,-5.64),dotstyle); label("$C$", (19.774908043768193,-5.2887524556302585), NE * labelscalefactor); dot((7.165969753513003,-1.3285130495083106),linewidth(4.pt) + dotstyle); label("$I$", (7.118709194975288,-3.087674394970628), NE * labelscalefactor); dot((-6.243576487131332,-5.753899126456025),linewidth(4.pt) + dotstyle); label("$D$", (-6.741204218240854,-4.5665237169763175), NE * labelscalefactor); dot((2.202200310840074,-2.96664425742365),linewidth(4.pt) + dotstyle); label("$E$", (1.1689200622547113,-2.7781477926903677), NE * labelscalefactor); dot((12.12973919618593,0.3096181584070275),linewidth(4.pt) + dotstyle); label("$F$", (12.277485899646308,0.5922529876946916), NE * labelscalefactor); dot((0.8133189006186733,0.10231412718850881),linewidth(4.pt) + dotstyle); label("$P$", (-0.06918634686633351,0.4202937642056579), NE * labelscalefactor); dot((10.576325071104362,-11.662372408466327),linewidth(4.pt) + dotstyle); label("$N$", (11.7272163844814,-11.238541588350822), NE * labelscalefactor); dot((10.55,-5.68),linewidth(4.pt) + dotstyle); label("$M$", (11.7272163844814,-6.630034398844721), NE * labelscalefactor); dot((5.629779458335351,-5.701651135496874),linewidth(4.pt) + dotstyle); label("$Q$", (5.502292494178368,-7.214695758707435), NE * labelscalefactor); dot((4.309242492561626,-9.460879739602213),linewidth(4.pt) + dotstyle); label("$T$", (3.679524725194608,-10.860231296674948), NE * labelscalefactor); dot((10.489221436718749,8.131928505665023),linewidth(4.pt) + dotstyle); label("$Y$", (11.589649005690173,7.1610953249757765), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] from $\angle DPN = \angle DIN = \angle DMN = 90 \implies D,P,I,M,N$ are cyclic. which give us $\angle IPM = \angle IDM$ Now as $I$ and $M$ are midpoint of $EF$ and $BC$ and $P$ lie on $(ABC)$ $P$ is miqual point of complete quadrilateral $BCFE$ and hence from $A = BE \cap CF$ we get $P$ lie on $(AEF)$ $\blacksquare$

This question is basically like a lemma

Take $IR \cap (ABC)$ at $R$.$\angle PAF=\angle PEF=\angle PBD$ so, $PEBD$ is cyclic $\implies$ $\angle EPB=\angle EDB=\angle FPC$ so, $PFCD$ cyclic. $\angle QPD=90$, $\angle BPQ=\angle QPC$ that means $(D,Q;C,B)=-1$.Take pencil from $P$ w.r.t $(ABC)$ so, $PBNC$ is harmonic quadrilateral.Then, take pencil from $N$ w.r.t line $BD$. So, $B$ fixes $B$, $P$ fixes $Q$ , $C$ fixes $C$ and $N$ fixes $NR \cap BD$ (call it $D'$). Thus, $(B,C;Q,D')=-1$ also $(B,C;Q,D)=-1$ and that means $D'=D$.

Let $T$ be the $A$ mixtilinear touch point. It is well known that $DN \cap \omega=T$. We prove that $\overline{T-Q-I}$. Since $(PT;BC) \stackrel{Q}{=} (N,TQ \cap \omega ;C,B)=-1$ $\implies TQ \cap \omega =M$ where $M$ is antipode of $N$ in $\omega$. But $I$ lies on $TM$ and so the required is proved.

Let $T_A$ the A-mixtilinear intouch point of $\triangle ABC$, by $\sqrt{bc}$ invert we have $M,I,T_A$ colinear, where $M$ is midpoint of arc $BAC$ in $\omega$. Let $L$ midpoint of $BC$. Claim 1: $D,P,M$ are colinear. Proof: Note that $P$ is miquel of $BEFC$ so by homologue points we have it's miquel of $BEIL$ as well so $DPIL$ is cyclic, but clearly $DILN$ is cyclic with diameter $DN$ therefore both combined give $\angle DPN=90$ which concludes. Finish: Note that $-1=(M, N; B, C) \overset{P}{=} (D, Q; B, C)$ and also that $DI$ is trivially tangent to $(BIC)$ from I-E Lemma, which means that in $\triangle BIC$, $IQ$ is the I-symedian but from trivial angle case we have $BN, NC$ tangent to $(BIC)$ therefore $N,I,Q,T_A$ are colinear, now projecting from $T_A$ gives $D,T_A,N$ colinear which finishes as $\angle NT_AM=90$ thus we are done .

I have not done mixtilinear circles or whatever, so heres a solution based purely on angle chase: Let $DN$ meet $\omega$ again in $L$. Let $IL$ meet $BC$ in $Q*$ and $QN$ meet $\omega$ in $P*$. We claim that $P$ and $P*$ are the same by showing that $AFEP$ is cyclic. Note that $\angle EDB=180-\angle EBD-\angle DEB=\frac{B-C}{2}$. Also, note that $\angle DIB=180-\angle IDB-\angle DBI= \frac{B-C}{2}$. Hence, $DB$ is tangent to $(BIC)$ at $I$. $\implies DI^2=DB \cdot DC=DL \cdot DN$ $\implies DI$ is tangent to $(LIN)$ in $I$. $\implies \angle DIN= \angle ILN=90$ $\implies \angle ILD=90$ Also note that $\angle LDQ=\angle BDL=180-\angle DBL-\angle DLB=180-\angle LNC-\angle BCN=180-\angle LNI-\angle B-\angle \frac{A}{2}=\angle C+\angle \frac {A}{2}-\angle LNI=\angle BCN +\angle ACB-\angle LNI=\angle ACN-\angle LNI=\angle ALD-\angle ANL=\angle LAN=\angle LP*N=\angle LP*Q*$, giving us the fact that $LDP*Q*$ is cyclic. $\implies \angle NP*D=\angle Q*P*D=\angle Q*LD=90$ $\implies \angle DP*C=\angle DP*N +\angle NP*C=90+\angle NAC=90+\angle \frac{A}{2}=180-(90-\angle \frac{A}{2})=\angle DFC$ $\implies DP*FC$ is cyclic $\implies \angle P*AC=\angle P*BA=\angle P*DF$ $\implies \angle P*DE=\angle P*BE$ $\implies EP*DB$ is cyclic $\implies \angle P*EA=\angle P*DB=\angle P*DC=\angle P*FA$, hence $P* \in (AEF)$, as required

Om245 wrote: We redefines everything Let $T$ be mixtilinear Incircles touch point.Note that circle pass though $E$ and $F$ Define $Y$ be antipode of $N$ then we know $\overline{Y-I-T}$ and $\angle YTN = 90$ Let $D = NT \cap BC$ and $M$ be midpoint of $BC$. Redefine $Q = DM \cap YT$ Now note $Q$ is orthocenter of $\triangle YND$. Redefine $P = DY \cap (ABC)$ now we get $NP \perp DY$ Now if we prove $P$ lie on $(AEF)$ we are done [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.603703365062785cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.889434907857462, xmax = 42.71426845720532, ymin = -19.217449558241984, ymax = 11.666426980388458; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw((4.676888221431007,6.213756330118006)--(1.46,-5.72)--(19.64,-5.64)--cycle, linewidth(0.8) + zzttqq); draw((10.489221436718749,8.131928505665023)--(-6.243576487131332,-5.753899126456025)--(10.576325071104362,-11.662372408466327)--cycle, linewidth(0.4) + qqwuqq); 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dot((0.8133189006186733,0.10231412718850881),linewidth(4.pt) + dotstyle); label("$P$", (-0.06918634686633351,0.4202937642056579), NE * labelscalefactor); dot((10.576325071104362,-11.662372408466327),linewidth(4.pt) + dotstyle); label("$N$", (11.7272163844814,-11.238541588350822), NE * labelscalefactor); dot((10.55,-5.68),linewidth(4.pt) + dotstyle); label("$M$", (11.7272163844814,-6.630034398844721), NE * labelscalefactor); dot((5.629779458335351,-5.701651135496874),linewidth(4.pt) + dotstyle); label("$Q$", (5.502292494178368,-7.214695758707435), NE * labelscalefactor); dot((4.309242492561626,-9.460879739602213),linewidth(4.pt) + dotstyle); label("$T$", (3.679524725194608,-10.860231296674948), NE * labelscalefactor); dot((10.489221436718749,8.131928505665023),linewidth(4.pt) + dotstyle); label("$Y$", (11.589649005690173,7.1610953249757765), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] from $\angle DPN = \angle DIN = \angle DMN = 90 \implies D,P,I,M,N$ are cyclic. which give us $\angle IPM = \angle IDM$ Now as $I$ and $M$ are midpoint of $EF$ and $BC$ and $P$ lie on $(ABC)$ $P$ is miqual point of complete quadrilateral $BCFE$ and hence from $A = BE \cap CF$ we get $P$ lie on $(AEF)$ $\blacksquare$ i'm fanumtaxing this diagram: Note that $D,P,Y$ collinear because $\angle DPC=\angle DFC$, $\angle CPD=\angle CNM$. This is easy to check as they are $90^\circ-\frac{\angle A}{2}$. Where we note that $DPFC$ is cyclic by Miquel. Suppose that $IY\cap (ABC)=T'$. We claim that $D,T',N$ collinear. This is radax on $(IT'N)$, $(BIC)$, and $(ABC)$. Now, note that $T'=T$. Note that as $PN\perp DY$, and $YT\perp DN$, and $DQ\perp YN$. Thus $Q$ also lies on $IT$.