Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent. Proposed by Alireza Dadgarnia
Problem
Source: Iranian TST 2020, second exam day 2, problem 4
Tags: geometry, incenter, circumcircle, Harmonics, homothety, Hi
12.03.2020 15:28
Nice and easy. Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$. Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$). Now since $$\angle NCH=\angle NHC=\angle ABC$$$\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$. But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$. Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$. But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$. Done $\blacksquare$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -10.006881897432205, xmax = 25.634412731501875, ymin = -8.577795838543958, ymax = 14.83226710342647; /* image dimensions */ pen ffqqff = rgb(1,0,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((1.5534973808986277,9.160392608982256)--(-3.5857910956748746,-1.8868970241407366), linewidth(0.8)); draw((1.5534973808986277,9.160392608982256)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); draw((-3.5857910956748746,-1.8868970241407366)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); draw((1.5534973808986277,9.160392608982256)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); draw(circle((5.807241390273251,2.031588419358404), 3.8827498197705226), linewidth(0.4) + wvvxds); draw(circle((1.9067865071150452,2.276975553541132), 6.892477317034895), linewidth(0.4) + dtsfsf); draw(circle((3.5081332893469326,4.891548128823499), 4.6950649335634775), linewidth(0.4) + wvvxds); draw((8.19899262334367,5.090227365365952)--(5.807241390273251,2.031588419358404), linewidth(0.4) + qqffff); draw((3.8176841460738786,5.365866172483034)--(8.19899262334367,5.090227365365952), linewidth(0.4) + qqffff); draw((8.19899262334367,5.090227365365952)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + wrwrwr); draw((3.8176841460738786,5.365866172483034)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); draw((1.9295955015644004,1.832569513956508)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + qqffff); draw((3.8176841460738786,5.365866172483034)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); draw((5.807241390273251,2.031588419358404)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); draw((1.9295955015644004,1.832569513956508)--(2.2600756333314593,-4.606441501899994), linewidth(0.4) + ffqqff); draw((2.2600756333314593,-4.606441501899994)--(8.19899262334367,5.090227365365952), linewidth(0.4) + green); draw((-1.0161468573881236,3.63674779242076)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); draw((1.5534973808986277,9.160392608982256)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); draw((1.9295955015644004,1.832569513956508)--(5.807241390273251,2.031588419358404), linewidth(0.4) + wrwrwr); draw((7.796798634472622,-1.3026893337662244)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); /* dots and labels */ dot((1.5534973808986277,9.160392608982256),dotstyle); label("$A$", (1.6396828828887742,9.383171533517269), NE * labelscalefactor); dot((-3.5857910956748746,-1.8868970241407366),dotstyle); label("$B$", (-3.4819992193005715,-1.6553396638953617), NE * labelscalefactor); dot((7.796798634472622,-1.3026893337662244),dotstyle); label("$C$", (7.88392544583195,-1.0706727572527435), NE * labelscalefactor); dot((1.9295955015644004,1.832569513956508),linewidth(4pt) + dotstyle); label("$I$", (2.0138697031400508,2.01636850982028), NE * labelscalefactor); dot((5.807241390273251,2.031588419358404),linewidth(4pt) + dotstyle); label("$N$", (5.896057963247044,2.2268485962116227), NE * labelscalefactor); dot((3.8176841460738786,5.365866172483034),linewidth(4pt) + dotstyle); label("$Q$", (3.9081904806621375,5.547756625941694), NE * labelscalefactor); dot((8.19899262334367,5.090227365365952),linewidth(4pt) + dotstyle); label("D", (8.28149894234893,5.267116510753237), NE * labelscalefactor); dot((-1.0161468573881236,3.63674779242076),linewidth(4pt) + dotstyle); label("$M$", (-0.9328515063387511,3.8171425822795437), NE * labelscalefactor); dot((17.798762343332783,-0.7893418218123583),linewidth(4pt) + dotstyle); label("Q", (17.893422887553594,-0.602939231938649), NE * labelscalefactor); dot((4.675148007685625,3.928851637608016),linewidth(4pt) + dotstyle); label("$F$", (4.773497502493214,4.121169373733705), NE * labelscalefactor); dot((4.169500681742824,-1.488859213336944),linewidth(4pt) + dotstyle); label("$H$", (4.258990624647709,-1.3045395199097907), NE * labelscalefactor); dot((2.2600756333314593,-4.606441501899994),linewidth(4pt) + dotstyle); label("$J$", (2.3646698471256222,-4.414967463248519), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
27.03.2020 00:56
CLAIM: $N$ is the centre of $\omega=\odot(CIQ)$. Proof: If $N'$ is the centre of $\omega$ then $N'C=N'I$ and $\angle N'IA=90^\circ$ implies $N'I$ and $BC$ are parallel .Let $IN'$ cut $AC$ at $N''$.Then $$\angle N''IC=\angle ICB=\angle ICN''$$.It means $N''C=N''I$.So $N''\equiv N'$.Now the circle centre at $N''$ and with radius $N''C=N''I$ is the unique circle tangent to $AI$ at $I$.So $N''=N$.. $\square$ Now let $\omega=\odot(CIQ)$ cut $BC$ at $X$ then $NX=NC$ implies $\angle NXH=\angle NXC=\angle ABC$ so $NX$ and $AB$ are parallel.So the line joining $C$ and the intersection of $BN$ and $AX$ passes through midpoint of $AB$.In other words $AX$,$BN$,$CM$ are concurrent at a point.So let $F$ is the intersection of $MN$ and $BC$.So $(B,C;X,F)=-1$ and let M be the midpoint of $BC$.Then $FC.FB=FX.FM$ [since $F$ is the inverse image of $X$ w.r.t circle with diameter $BC$]. Now assume $Y= XN \cap AD$.We claim that $XYDC$ is cyclic.Indeed, as $XN$ and $AB$ are parallels so $\angle XYD=\angle BAD=180^\circ-\angle XCD$. Now as $XY$ passes through center $N$ of $omega=\odot(CIQ)$ so $\angle ADX=\angle XDY=90^\circ$ and together with $\angle AMXY=90^\circ$ We get $AMXD$ is cyclic. Let $F'= BC\cap AD$.So $F'D.F'A=F'C.F'B$ and $F'D.F'A=F'X.F'M$. So we have $F'C.F'B=F'X.F'M$.But we have previously proved that $FC.FB=FX.FM$ so $F \equiv F'$. So $BC,AD,MN$ concur at a point $F$.
12.04.2020 11:48
30.06.2020 15:24
17.05.2021 23:37
Claim: N is the circumcenter of (CIQ) Proof: Note that $IN\perp, AI, BC\perp AI \Longrightarrow NI\parallel BC$. Thus, \[\angle NIC = \angle BCI = \angle ICA = \angle ICN\]Thus, $NI=NC$, so the circle with center $N$ and radius $NI$ is both tangent to $AI$ at $I$ and passes through $C$, so we have sufficiently redefined $\omega$. $\square$ Now, define $X=\omega \cap BC$ and $Y=\omega \cap AD$ Claim 1: $NX\parallel AB$ This clearly follows from \[\angle CNX = \angle NCX = \angle ACB=\angle ABC\]$\square$ Claim 2: $XY\parallel AB$ We angle chase (basically a rederivation of Reim's) \[\angle DAB = \angle DCB = \angle DCX = \angle DYX\]$\square$ Combining these, we have that $N,X,Y$ are collinear. Since, $X,Y\in \omega$ and $N$ is the center of $\omega$, we have that $N$ is the midpoint of $XY$. Since $M$ is the midpoint of $AB$, combined with $YX\parallel AB$, we have that $AY,MN,BX$ are concurrent which finishes $\blacksquare$
13.08.2021 18:04
Attachments:

13.08.2021 19:35
$\angle QIA = \angle QCI = \angle ICB = 90^\circ - \angle (IC , \text{angle bisector}) \implies \angle QIC = 90^\circ \implies N$ is the circumcenter of $\omega$. Let $E=BC \cap \omega$ and $F=AD \cap \omega$. $$\angle FEC = 180^\circ - \angle FDC = \angle ABC = \angle NCE = \angle NEC$$Thus $\overline{E-N-F}$ are collinear also $EF \parallel AB$ so by homothety we are done.
25.09.2021 20:05
To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles. Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$
09.11.2021 21:13
Consider $P$ as $AI \cap BC$. Claim $N$ is the centre of the circle $\omega$ Proof We angle chase. Note that since $ \angle IQC = \angle PIC = 90^{\circ} - \angle ICP = 90^{\circ} - \angle C/2 $ and $\angle QCI = \angle C/2$ We have that $\angle QIC = 90^{\circ}$ and thus $QC$ is a diameter But since $N$ is the midpoint of $QC$ it must be the centre of the circle We now define $X$ as $AD\cap \omega (\neq D)$ and $Y$ as $BC\cap \omega (\neq C)$ Note that $\angle NXC = \angle NCX$ but $\angle NCX = \angle ABC$ since the triangle is isosceles. Thus $\angle NXC = \angle ABC$ and so $NX||AB$ However we also have that $XY||AB$ by Reim's Theorem. Combining these we have that $ABXY$ is a trapezium and $M$ and $N$ are the midpoints of its parallel sides. A simple homothety argument proves that $AY, BX,MN$ that is $AD,BC,MN$ are concurrent.
06.12.2021 08:30
srisainandan6 wrote: To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles. Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$ Can you explain Reim's theorem please?
23.12.2021 18:34
14.02.2022 19:19
Here's a solution with radical axes ($\overline{MN}$ will become radical axes of two circles)
Claim: Points $M,N,T$ lie on radical axes $\ell$ of $\Gamma,\gamma$. Proof: $T \in \ell$ is direct. For $M$: Note $\Gamma$ is tangent to $\overline{BC}$ as it has center $A'$. Note $$\angle OAC = \angle OCA = \angle C'CA = \angle C'DA = \angle QDA$$As $\angle BAC = 2 \angle OAC = 2 \angle QDA = \angle XDA$, so $\gamma$ is tangent to $\overline{AB}$. Hence power of $M$ wrt both $\Gamma,\gamma$ equals $MB^2 = MA^2$. For $N$: We will mostly focus of $\triangle ADX$. Observe $$ \angle NDX = \angle NDQ - \angle XDQ = \angle NQD - \angle ADQ = \angle QAD = \angle NAD $$So $\overline{ND}$ is tangent to $\gamma$ (experts may also directly note that $N$ is the center of $D$-appolonius circle wrt $\triangle ADX$). Hence $NX \cdot NA = ND^2 = NC^2$. Laslty, recall $\overline{NC}$ is tangent to $\Gamma$ (as $A'$ is its center). This proves our Claim. $\square$ It follows points $M, N,T$ are collinear, solving our problem. $\blacksquare$
07.03.2022 11:24
Simple angle chasing gives us that $N$ is the center of $(IDC)$. Let $AD\cap BC=S$. We have $$\angle ASC=\angle ACB-\angle CAD=\angle ABC-\angle CAD=\angle ACD$$Hence, $NC$ is tangent to $(DCS)$. Since $|NC|=|ND|$, we know that $ND$ is tangent to $(DCS)$ as well. Then, $SN$ is symmedian in $DSC$. Since $DC$ and $AB$ are antiparallels wrt $DSC$, we find that $SN$ bisects $AB$, done.
07.03.2022 11:36
07.03.2022 13:27
BarisKoyuncu wrote:
Proof. All angles are oriented. From $\angle ADB=\angle ACB=\angle CBA=\angle SBA$ we obtain $$\angle CDN=\angle NCD=\angle ABD=\angle CBS.$$Therefore $SN$ is a symmedian in $CDS$ and so bisects $AB.$
07.03.2022 14:09
GeoMetrix wrote: Nice and easy. Define $\omega=\odot(CIQ)$ and let $H=\omega \cap \overline{BC}$. Observe that trivially we have that $N \in \overline{AC}$ (just consider the line through $I$ parallel to $BC$). Now since $$\angle NCH=\angle NHC=\angle ABC$$$\implies$ $\overline{NH} \| \overline{AB}$ . Hence by the converse of ceva's theorem we have that $\overline{AH},\overline{BN},\overline{CM}$ are concurrent. Now define $K= \overline{AD} \cap \overline{BC}$ . Observe that to show that $K \in \overline{MN}$ we need to show that $(B,C;H,K)=-1$. But if $\overline{DH} \cap \omega=J$ then we have that $(B,C;H,K)\overset{D}{=}(A,J;B,C)$. Hence we just need to show that $(AJBC)$ is harmonic. For this notice that by reims theorem we have that $\overline{AJ} \| \overline{QH}$. But as $\angle QHC=90^\circ$ we have that $\overline{AJ} \perp \overline{BC}$ by which we have that $J$ is the midpoint of minor $\widehat{BC}$. Done $\blacksquare$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -10.006881897432205, xmax = 25.634412731501875, ymin = -8.577795838543958, ymax = 14.83226710342647; /* image dimensions */ pen ffqqff = rgb(1,0,1); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((1.5534973808986277,9.160392608982256)--(-3.5857910956748746,-1.8868970241407366), linewidth(0.8)); draw((1.5534973808986277,9.160392608982256)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); draw((-3.5857910956748746,-1.8868970241407366)--(7.796798634472622,-1.3026893337662244), linewidth(0.8)); draw((1.5534973808986277,9.160392608982256)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); draw(circle((5.807241390273251,2.031588419358404), 3.8827498197705226), linewidth(0.4) + wvvxds); draw(circle((1.9067865071150452,2.276975553541132), 6.892477317034895), linewidth(0.4) + dtsfsf); draw(circle((3.5081332893469326,4.891548128823499), 4.6950649335634775), linewidth(0.4) + wvvxds); draw((8.19899262334367,5.090227365365952)--(5.807241390273251,2.031588419358404), linewidth(0.4) + qqffff); draw((3.8176841460738786,5.365866172483034)--(8.19899262334367,5.090227365365952), linewidth(0.4) + qqffff); draw((8.19899262334367,5.090227365365952)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + wrwrwr); draw((3.8176841460738786,5.365866172483034)--(1.9295955015644004,1.832569513956508), linewidth(0.4) + ffqqff); draw((1.9295955015644004,1.832569513956508)--(7.796798634472622,-1.3026893337662244), linewidth(0.4) + qqffff); draw((3.8176841460738786,5.365866172483034)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); draw((5.807241390273251,2.031588419358404)--(4.169500681742824,-1.488859213336944), linewidth(0.4) + qqffff); draw((1.9295955015644004,1.832569513956508)--(2.2600756333314593,-4.606441501899994), linewidth(0.4) + ffqqff); draw((2.2600756333314593,-4.606441501899994)--(8.19899262334367,5.090227365365952), linewidth(0.4) + green); draw((-1.0161468573881236,3.63674779242076)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); draw((1.5534973808986277,9.160392608982256)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); draw((1.9295955015644004,1.832569513956508)--(5.807241390273251,2.031588419358404), linewidth(0.4) + wrwrwr); draw((7.796798634472622,-1.3026893337662244)--(17.798762343332783,-0.7893418218123583), linewidth(0.4) + linetype("4 4") + blue); /* dots and labels */ dot((1.5534973808986277,9.160392608982256),dotstyle); label("$A$", (1.6396828828887742,9.383171533517269), NE * labelscalefactor); dot((-3.5857910956748746,-1.8868970241407366),dotstyle); label("$B$", (-3.4819992193005715,-1.6553396638953617), NE * labelscalefactor); dot((7.796798634472622,-1.3026893337662244),dotstyle); label("$C$", (7.88392544583195,-1.0706727572527435), NE * labelscalefactor); dot((1.9295955015644004,1.832569513956508),linewidth(4pt) + dotstyle); label("$I$", (2.0138697031400508,2.01636850982028), NE * labelscalefactor); dot((5.807241390273251,2.031588419358404),linewidth(4pt) + dotstyle); label("$N$", (5.896057963247044,2.2268485962116227), NE * labelscalefactor); dot((3.8176841460738786,5.365866172483034),linewidth(4pt) + dotstyle); label("$Q$", (3.9081904806621375,5.547756625941694), NE * labelscalefactor); dot((8.19899262334367,5.090227365365952),linewidth(4pt) + dotstyle); label("D", (8.28149894234893,5.267116510753237), NE * labelscalefactor); dot((-1.0161468573881236,3.63674779242076),linewidth(4pt) + dotstyle); label("$M$", (-0.9328515063387511,3.8171425822795437), NE * labelscalefactor); dot((17.798762343332783,-0.7893418218123583),linewidth(4pt) + dotstyle); label("Q", (17.893422887553594,-0.602939231938649), NE * labelscalefactor); dot((4.675148007685625,3.928851637608016),linewidth(4pt) + dotstyle); label("$F$", (4.773497502493214,4.121169373733705), NE * labelscalefactor); dot((4.169500681742824,-1.488859213336944),linewidth(4pt) + dotstyle); label("$H$", (4.258990624647709,-1.3045395199097907), NE * labelscalefactor); dot((2.2600756333314593,-4.606441501899994),linewidth(4pt) + dotstyle); label("$J$", (2.3646698471256222,-4.414967463248519), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] When you use Geogebra-Asymptote conversion, how did you doit. Why did mine have an error?
07.03.2022 14:31
Firstly you have convert it from Geogebra Classic and not Geogebra Geometry. If you find easier to draw in Geogebra Geomtery (as I also do), then save it and reopen it in Geogebra Classic. After the conversion, you have to do some slight edits. Like changing the size and defaultpen.
14.04.2022 19:02
We begin with the following key claim. Claim: $N$ is the center of $\omega$. Proof: Let $N'$ be the intersection of the perpendicular to $\overline{AI}$ at $I$ with $\overline{AC}$. Then we have $$\angle N'IC=\angle ICB=\angle ICA=\angle ICN',$$so $N'I=N'C$ and thus $N'$ is the center of $\omega$. Then $Q$ is the reflection of $C$ over $N'$, hence $N'$ is the midpoint of $\overline{CQ}$ and we have $N=N'$, which implies the desired result. Now let $P \neq D$ be the intersection of $\omega$ with $\overline{AD}$, $R \neq C$ be the intersection of $\omega$ with $\overline{BC}$, and $X=\overline{AD} \cap \overline{BC}$. By Reim's, we have $\overline{AB} \parallel \overline{PR}$, so $$\angle QPR=\angle QCR=\angle ABC=\angle PRC,$$so we have $\overline{PQ} \parallel \overline{CR}$ as well. Then we have $90^\circ=\angle QPC=\angle PCR$, so $CPQR$ is a rectangle, and $N$ is the midpoint of $\overline{PR}$. But triangles $\triangle XAB$ and $\triangle XPR$ are homothetic, which implies that $\overline{MN}$ passes through $X$ as well, hence $\overline{AD}$, $\overline{MN}$, and $\overline{BC}$ concur. $\blacksquare$
05.06.2023 18:37
mohamad021 wrote: srisainandan6 wrote: To begin, note that it is easy to see that $N$ is the circumcenter of $\omega$ as $\triangle NIC$ is isosceles. Denote $X= BC \cap AD$. We desire to show that $M,N,X$ are collinear. Denote $E = \omega \cap AD$ and $F = \omega \cap BC$. By Reim's Theorem, we get that $EF \parallel AB$. To finish, take a homotethy centered at $X$ that maps $\triangle XEF$ to $\triangle XAB$ which yields $X,N,M$ collinear. $\blacksquare$ Can you explain Reim's theorem please? JI//FH,as the picture says
Attachments:

08.09.2023 06:03
By angle chasing, $\angle QIC = 90^{\circ}$, hence $\angle QDC = 90^{\circ}$, hence $\angle ADQ = 90 - \angle B$, hence $(ADQ)$ is tangent to $\overline{AI}$. If we extend $AB$ to $B'$ and $AC$ to $C'$ such that $BB' = CC' = AQ$, then $M$ and $N$ have equal powers WRT $(AQD)$ and $(BCC'B')$, hence $\overline{MN}$ is their radical axis. Then, obviously $\overline{BC}$ is the radical axis of $(ABC)$ and $(BCC'B')$, and $\overline{AD}$ is the radical axis of $(AQD)$ and $(ABC)$, hence they all concur by the radical axis theorem.
16.10.2023 21:45
Claim: $N$ is the center of $(CIQ)$ Proof: By angle-chasing, we can see that: \[\angle CQI=\angle CIF=90^{\circ}-\angle ICF=90^{\circ}-\angle ICQ.\]Therefore, $\angle QIC=90^{\circ}$, or $N$ is the center $\blacksquare$ Claim: The line parallel to $AB$ that passes through $N$ goes through $E$ and $F$. Proof: Denote $E=(CQD)\cap BC$ and $F=(CQD)\cap AD$. By Reim's theorem, $EF$ is parallel to $AB$. Also, $\triangle{NEC}$ is isosceles, so it is similar to $\triangle{ABC}$. As a result, $NE\parallel AB$. Therefore, $N,E,F$ are all collinear and parallel to $AB$ $\blacksquare$ As a result, we can take a homothety at $T=AD\cap BC$, which sends $N$ to $M$, implying the desired result $\square$
29.10.2023 04:54
Note that $N$ is the center of $\omega$. Let $E$ be the other intersection of $\omega$ with $BC$ and let $E'$ be its antipode wrt $\omega$. Let $F$ be the midpoint of $BC$. It suffices to show that $AFED$ is cyclic as this implies $A,E',D$ collinear. Here's the interesting part. Claim: Let $ABCD$ be a cyclic quadrilateral. Let $F$ and $E$ be points where $B,F,E,C$ are on $BC$ in that order. If $\angle BAF=\angle CDE$ then $AFED$ is cyclic. Proof: Angle chase. Now we're done; just apply the claim here.
24.11.2023 17:32
Claim:$N$ is the center of $\omega$. Proof Let $O$ denote the center of $\omega$. $\angle ICQ= \angle ICB = \tfrac{1}{2} \angle QOI$. Since $OI \perp AI$, then $OI \parallel BC$. Since $\angle QOI = \angle ICQ + \angle ICB= \angle BCA$, $O$ lies on $AC$. Since $OQ=OC$, then $O$ is the midpoint of $CQ$. So $O= N$. $\square$ By Reims on $\omega$ and $(ABC)$ with lines $AD$ and $BC$ we get that the line through $D$ parallel to $AB$ intersects $BC$ on $\omega$ at a point we will call $P$ then let $DP\cap MN= Q$ then by homothety $BP$, $AD$, $MQ$ are concurrent so $MN$, $BC$, and $AD$ are concurrent.
24.11.2023 20:01
Define $R$ as midpoint of $BC$ Easy to see $N$ is center of $DQIC$. After that define $S=(DIQ) \cap BC$ Angle chasing shows $SN//AB$ so $AS,BN,CM$ concurrent. Now we should show that $(B,C;S,AD \cap BC)=-1$ Let $AD \cap BC= K'$ so $(B,C;S,AD \cap BC)=-1$ is equal to $KD.KA=KS.KR$ and from angle chasing it is easy that show $ADSR$ is cyclic.
24.12.2023 00:18
If $N'$ is the intersection of $AC$ with the line through $I$ parallel $BC$, we know $N'I \perp AI$ and \[\angle ACI = \angle ICB = \angle NIC,\] so $N'$ is the center of $\omega$. Therefore $N'$ is the midpoint of $CQ$, so $N' = N$. Next we define the intersections of $AC$ and $BC$ with $\omega$ as $K$ and $L$. Looking at $CD$, Reim's tells us that $AB \parallel KL$, which then gives \[\angle LKQ = \angle LCQ = \angle ABC = \angle KLC,\] so $KQ \parallel BC$ as well. As a result, $\angle KCL = \angle QKC = \angle QLC = 90$, so $KL$ is a diameter of $\omega$, and hence passes through $M$. We finish by noting the homothety which maps $KL$ to $AB$ also maps corresponding midpoints $N$ to $M$.