Solution 1(Rohan Goyal) Let $I$ and $I_A$ denote the incentre and excentre opposite to $A$ of scalene $\triangle ABC$ respectively. Let $A'$ be the antipode of $A$ in $\odot (ABC)$ and $L$ be the midpoint of arc $(BAC)$. Let $LB$ and $LC$ intersect $AI$ at points $Y$ and $Z$ respectively. Prove that $\odot (LYZ)$ is tangent to $\odot (A'II_A)$. Solution. WLOG $AB<AC$. Let $D=AI\cap BC, L'=AI\cap (ABC), P=DL'\cap (ABC), X=DL'\cap \ell$ where $\ell$ is tangent to $(ABC)$ at $L$. $A'I\cap BC = R$. Claim 1. $BDPY$ is cyclic and similarly $CDPZ$ cyclic. Proof. $\angle BYD=180-(C+\frac{A}{2})-(90-\frac{A}{2})=90-C=\angle BAA'=\angle BPA'=\angle BPD$.$\square$ Claim 2. $PYLX$ is cyclic and thus similarly $PZLX$ is cyclic and thus $(LYZ)=(PYZLX)$. Proof. $\angle PYB=\angle PDB = \angle XDB=\angle LXD=\angle LXP$. $\square$ Claim 3. $PIA'I_A$ is cyclic. Proof. $DI\cdot DI_A= DB\cdot DC=DA'\cdot DI_A$.$\square$ Claim 4. $A'I=A'I_A$. Proof. $A'L'\perp II_A$ but $L'I=L'I_A$. Claim 5. $\angle A'PL=\angle A'I_AP+\angle PXL$ Proof. $\angle PXL+\angle A'I_AP =\angle PDB+\angle II_AP+\angle A'I_AI=\angle A'DR+\angle RA'D+\angle I_AIA'=\angle IRD+\angle DIR=\angle IDB=\angle ADB=C+\frac{A}{2}=\angle A'PL$ Now, let $T$ be any point on the tangent at $P$ to $(A'II_A)$ inside $\angle A'PL$. Now, $\angle A'PL=\angle A'I_AP+\angle PXL=\angle PXL+\angle A'PT=\angle PXL +\angle A'PL - \angle TPL\implies \angle PXL=\angle TPL\implies TP$ is tangent to $(PXL)$ i.e. $(LYZ)$. Thus, $TP$ is tangent to $(A'II_A)$ and $(LYZ)$ at $P$. Thus, they are both tangent to each other. $\square$ Solution 2 (Mahavir Gandhi) Let $L'$ be midpoint of arc $BC$ not containing $A$ in $\odot (ABC)$ and let $AD$ be altitude of $\triangle ABC$ with $D\in BC$. Let $DL'$ intersect $\odot (ABC)$ again at point $P$. Let $AI\cap BA'=K$. Claim 1: $P\in\odot (A'II_A)$. Proof: Let $AL'\cap BC=K$. Then we have that $\triangle ADK\sim\triangle AL'A'$. Hence there exists spiral similarity with centre $A$ taking $DK$ to $L'A'$. Thus tehre exists spiral similarity with centre $A$ taking $DL'$ to $KA'$. So we have that $DL'$ and $KA'$ intersect on $\odot (AL'A')$ ie $\odot (ABC)$. Hence $P\in KA'$. Now consider cyclic quadrilaterals $BICI_A$ and $BPCA'$. Since $K\in BC, II_A, PA'$ we have $KP.KA'=KB.KC=KI.KI_A$. Hence $PIA'I_A$ is cyclic as desired. Claim 2: $P\in\odot (LYZ)$ Proof: Let $\odot (LYZ)$ and $\odot (ABC)$ intersect at point $P'\neq L$. Then we have that there exists spiral similarity with centre $P'$ taking $BY$ to $CZ$. Hence $\triangle P'BY\sim\triangle P'CZ$. Hence $\frac{}{}\frac{P'B}{P'C}=\frac{BY}{CZ}$. Now by applying sine rule in $\triangle ABY$ we have that $\frac{BY}{sin\left(\frac{A}{2}\right)}=\frac{AB}{sin(90^{\circ}+C)}$, ie $BY=\frac{ABsin\left(\frac{A}{2}\right)}{cos(C)}$ By applying sine rule in $\triangle ACZ$ we have that $\frac{CZ}{sin\left(180^{\circ}-\frac{A}{2}\right)}=\frac{AC}{sin(90^{\circ}-B)}$, ie $CZ=\frac{ACsin\left(\frac{A}{2}\right)}{cos(B)}$ Hence we have $\frac{P'B}{P'C}=\frac{BY}{CZ}=\frac{ABcos(B)}{ACcos(C)}=\frac{DB}{DC}$. Thus $P'D$ bisects $\angle BP'C$ or in other words $P',D,L'$ are collinear. Hence $P'\equiv P$. Thus we have $P\in\odot (LYZ)$ as desired. Claim 3: $\odot (LYZ)$ and $\odot (A'II_A)$ are tangent at $P$. Proof: These circles are tangent iff $\angle PLY+\angle PA'I=\angle YPI$. But $\angle PLY+\angle PA'I=\angle PLB+\angle PA'I=\angle PA'B+\angle PA'I=\angle BA'I=\angle KA'I$. $\angle YPI=\angle LPA'-\angle LPY-\angle IPA'=\left(C+\frac{A}{2}\right)-\angle LZY-\angle A'I_AI= \left(C+\frac{A}{2}\right)-(90^{\circ}-B)-\angle A'II_A=\left(90^{\circ}-\frac{A}{2}\right)-\angle A'IK=\angle AKB-\angle A'IK=\angle KA'I$. Thus $\angle PLY+\angle PA'I=\angle KA'I=\angle YPI$ as desired. This proves claim $3$ and hence the problem. Q.E.D

Solved with bora_olmez, i3435 Let $AI \cap BC = K$. Let $KA' \cap (ABC) = T$. We claim $T$ is the desired point. By PoP, we have $TK.KA' = BK.KC = KI.KI_A$ so $T \in (A'II_A)$. Note that since $\angle BTK = \angle BTA' = \angle BLA' = \angle LYA = \angle BYK$ so $BKYT$ cyclic and similarly $CKTZ$ cyclic. So, $\angle YTL = \angle LTA' - \angle YTA' = \angle LBA' - \angle LBC = \angle CBA' = \angle CLA' = \angle CZA = \angle LZY$ so $T \in (LYZ)$ too. For the tangency, we want $\angle LTA' = \angle LYT + \angle TI_AA'$. But $\angle LYT = \angle TZC = \angle TKC$ Extend $I_AA'$ and $IA'$ to meet $(ABC)$ at $S,R$. $R,S$ are symmetric wrt perpendicular bisector of $BC$ because they are the sharkydevil point and the excircle version of sharkydevil. So $\angle TI_AA' = \angle TI_AI + \angle AI_AS = \angle TA'R + \angle AI_AS = 90 - \angle RTA' + 90 - \angle MAS = 180 - \angle RTA' - \angle MAS = \angle RAA' - \angle MAS = \angle MAA'$ So $\angle LYT + \angle TI_AA' = \angle TKC + \angle MAA' = \angle LTA'$, so the circles are indeed tangent, as desired. $\blacksquare$

L567 wrote: Let $AI \cap BC = K$. Let $KA' \cap (ABC) = T$. We claim $T$ is the desired point. Can smone tell me how to come up with this point?

Let $E=\overline{AI}\cap\overline{BC}$, let $M=\overline{AI}\cap\odot(ABC)$ and $D=\overline{A'E}\cap \odot (ABC)$. I contend that $D$ is the desired tangency point. Observe that $D$ lies on $\odot (A'II_A)$ by trivial PoP. Claim: Let $F=\overline{LM}\cap\odot (LYZ)$. Let $D'=\overline{FA}\cap\odot (LYZ)$. Then, $D\equiv D'$. Proof. Firstly, I claim that $D'$ lies on $\odot (ABC)$. Indeed, as $\triangle FLY\sim\triangle FYM$ by trivial angle chase, by Shooting Lemma and PoP, \begin{align*} FA\cdot FD'=FY^2=FL\cdot FM, \end{align*}which means that $D'$ lies on $\odot (ABC)$. Now, observe that $D'$ is the Miquel point of complete quadrilateral $CEBYZL$ and by trivial angle chase again, $\measuredangle ED'A=90^\circ$, which means that $D'$ lies on $\overline{A'E}$. Thus, $D\equiv D'$. $\square$ Let $T$ be intersection of $\overline{AI}$ and tangent from $D$ to $\odot (LYZ)$. Observe that $TD=TA$, thus $T$ is the midpoint of $\overline{AE}$. Now, as $(A,E;I,I_A)=-1$, we get that $TE^2=TI\cdot TI_A$, hence all in all, $TD^2=TI\cdot TI_A$, thus $\overline{TD}$ is also tangent to $\odot(A'II_A)$. $\blacksquare$

SPHS1234 wrote: L567 wrote: Let $AI \cap BC = K$. Let $KA' \cap (ABC) = T$. We claim $T$ is the desired point. Can smone tell me how to come up with this point? We thought that the circles might intersect on $(ABC)$, and then Miquel/angle chase things would be of use. The radical axis of $(A'II_A)$ and $(ABC)$ goes through $K$ by radical axis with these circles and $(BIC)$.

Let $M_a$ be the midpoint of arc $BC$, the reflection of $A'$ over $M_a$ be $A_2$, $D = AA_2 \cap BC$, and $E = AM_a \cap BC$. Now, we consider an inversion about $(BICI_a)$, which is centered at $M_a$. It's easy to see that $Y^*, Z^*$ are the projections of $B, C$ onto $AI$ respectively. We also know that $L^*$ is the midpoint of $BC$ and $A'^* = M_aA' \cap BC$. Hence, it suffices to show $(L^*Y^*Z^*)$ and $(A'^*II_a)$ are tangent. The Shooting Lemma and Fact $5$ implies $$M_aA'^* \cdot M_aA_2 = M_aA'^* \cdot M_aA' = M_aB^2 = M_aI \cdot M_aI_a$$so $A_2$ lies on $(A'^*II_a)$. In addition, we have $M_aI = M_aI_a$ and $$\angle IM_aA'^* = \angle AM_aA' = 90^{\circ}$$so $A'^*A_2$ is actually a diameter of this circle. It's easy to see that $ALA'M_a$ is a rectangle. Thus, we know $$AL \parallel A'M_a \equiv A_2M_a$$and $$AL = A'M_a = A_2M_a$$so $ALM_aA_2$ is a parallelogram. Now, we have $$AA_2 \parallel LM_a \perp BC$$so $D$ is actually the projection of $A$ onto $BC$. This yields $\angle A'^*DA_2 = 90^{\circ}$, so $D$ also lies on $(A'^*IA_2I_a)$. Thales' implies $ACZ^*D$ is cyclic, and the Iran Lemma gives $L^*Y^* \parallel AC$. Thus, $Y^*L^*Z^*D$ is cyclic by Reim's. Now, we will show that $D$ is the desired tangency point. Once again, Iran Lemma yields $L^*Z^* \parallel AB$, and it's easy to see $A_2DEM_a$ is cyclic via Thales'. Hence, we have $$\angle DZ^*L^* = \angle DZ^*A + \angle AZ^*L^* = \angle DCA + \angle Z^*AB$$$$= \angle ECA + \angle EAC = \angle AED = \angle AA_2M_a = \angle DA_2A'^*$$which finishes since $D, L^*, A'^*$ are collinear. $\blacksquare$

This is my generalization problem. Generalization problem. Let $ABC$ be a triangle, $A'$ be the antipode of $A$ in $\odot (ABC)$. $P$, $Q$ are points such that $\angle BAP = \angle CAQ$ and $\angle PBQ = \angle PCQ = 90^\circ$. Line passes through $A$ and perpendicular to $PQ$ cuts $\odot (ABC)$ at second point $L$. $PQ$ cuts $LB, LC$ at $X,Y$, respectively. Prove that $\odot (LXY)$ is tangent to $\odot (A'PQ)$.

Attachments:

Really nice Problem Let $I_aA'I$ meet $ABC$ at $S$, $AI$ meet $BC$ at $K$ and $T$ be midpoint of arc $BC$. Claim1 : $A'I = A'I_a$. Proof : It's well-known that $T$ is midpoint of $II_a$ we also have $\angle A'TI = \angle A'TA = \angle 90$ so $A'$ lies on perpendicular bisector of $II_a$ so $A'I = A'I_a$. Claim2 : $A',K,S$ are collinear. Proof : Let $A'K$ meet $I_aA'I$ at $S'$. we have $S'K.KA' = IK.KI_a = BK.KC$ so $S'$ lies on $ABC$ so $S'$ is exact $S$. Claim3 : $SYKB$ is cyclic. Proof : $\angle BYK = \angle A/2 + \angle B/2 - \angle C/2 = \angle 90 - \angle C = \angle BAA' = \angle BSA' = \angle BSK$. Claim4 : $ZLYS$ is cyclic. Proof : $\angle YZL = \angle A/2 - \angle B/2 - \angle C/2 = (\angle 180 - \angle B/2 - \angle C/2) - (\angle B + \angle A/2) = \angle BSL - \angle BSY = \angle YSL$. Let $\ell$ be tangent to $I_aA'I$ at $S$ and $X$ an arbitrary point on it. Claim5 : $\ell$ is tangent to both $I_aA'I$ and $LYZ$. Proof : $\angle YSX = \angle YSK - \angle XSK = \angle B/2 + \angle C/2 - \angle XSI - \angle ISA' = \angle B/2 + \angle C/2 - \angle IKS = \angle SLB = \angle SLY = \angle SZY$ so $\ell$ is tangent to $LYZ$. we're Done.

I mean...its ok to get scared of this but an inversion will calm ur soul. (Wlog $AB \le AC$) Step 1: Identify the point of tangency $X$ Let $H$ be the projection of $A$ over $BC$, $AI$ hits $BC,(ABC)$ in $K,N$ respectivily, now let $(IA'I_A)$ hit $(ABC)$ again at $X$, by radax on $(ABC),(IA'I_A),(BICI_A)$ we get that $X,K,A'$ are colinear and that means $AXHK$ cyclic. Now by angle chase: $$\angle BYK=90-\angle ALB=\angle HAC=\angle BAA'=\angle BXK \implies BYXK \; \text{cyclic}$$Hence $X$ is the miquel point of $KYLC$ meaning that $CZXK$ and $ZXYL$ are cyclic. Step 2: Invert, no more words lol Make a $\sqrt{bc}$ inversion then $NA' \cap BC=X'$ and it holds that $(HIX'I_A)$ is cyclic and using the pervious cyclic quads we get that $Y',Z'$ are points in $AI$ such that $Z'ABL', Y'ACL', L'X'Y'Z'$ is cyclic, now by I-E lemma $IN=NI_A$ hence $NX'$ is the perpendicular bisector of $II_A$ and so, it passes through the center of $(HIX'I_A)$, also notice that $\angle Z'L'X'=\angle BAI=\angle CAI=\angle X'L'Y'$ so $X'$ is midpoint of arc $Z'Y'$ in $(L'Z'X'Y')$ and that means $NX'$ is the perpendicular bisector of $Y'Z'$ meaning that $NX'$ passes through the center of $(L'Z'X'Y')$ hence the centers of $(L'X'Y'Z'),(HIX'I_A)$ and $X'$ are colinear hence those circles are tangent, inverting back we get $(IA'I_A),(LYZ)$ tangent as desired. Thus we are done

IMHO this is > IMOP5 level. Will add soln later

Rename $Z$ as $X$.Let $R= (LXY) \cap (ABC)$ , $D = AI \cap BC$. $\angle XYL = \angle ADB - \angle LCB = 90 - \angle C$ and $\angle YXL = 180 - \angle ADC - \angle XBC = 90 - \angle B$ Claim : $\triangle RXY \sim \triangle RBC$ $\angle RXY = \angle RLY = \angle RBC$ , $\angle RYX = \angle RLX = \angle RCB$ and $\angle XRY = \angle BAC$ Hence Spiral Similarty At $R$ give us $XY \rightarrow BC$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15.68344575886152cm); 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 = -34.284782853597356, xmax = 87.39866290526416, ymin = -31.398995217119992, ymax = 32.72760941401721; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); 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label("$J$", (13.417413019007986,-20.40178239619), NE * labelscalefactor); dot((9.270477575842321,4.822737019189902),linewidth(4.pt) + dotstyle); label("$X$", (7.347522825637546,4.877524997376334), NE * labelscalefactor); dot((5.026996304164302,23.374871474270407),linewidth(4.pt) + dotstyle); label("$Y$", (3.848409655341645,21.15911281018177), NE * labelscalefactor); dot((17.10608522189399,-10.871097583622438),linewidth(4.pt) + dotstyle); label("$A'$", (17.41639949934616,-8.262002009449105), NE * labelscalefactor); dot((1.2909477684390025,0.9391878465041423),linewidth(4.pt) + dotstyle); label("$R$", (-2.078659592302432,-0.5496709402254782), NE * labelscalefactor); dot((11.985523767835376,-7.047210974168738),linewidth(4.pt) + dotstyle); label("$D$", (12.27484545319708,-6.476740187869562), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim : This Spiral Similarity also give us $\overline{RL} \rightarrow {RD}$ (that mean image of $L$ will lie on $RD$ only) Note some spiral similarity $B \rightarrow X$ and $C \rightarrow Y$.Now $$\angle BRX = 180 - \frac{\angle A}{2} - \angle C$$$$\angle LRD = \angle LRX + \angle XRD = \angle LYX + \angle XBL = 90 - \angle C + 90 - \frac{A}{2} = 180 - \frac{\angle A}{2} - \angle C$$ Let $L' = RD \cap (ABC)$.From our claims we get this similarity also send $L$ to $L'$ Note $\angle XRL = 90 - \angle C$ and $ \angle LRY = 90 - \angle B$. so we get $\angle BRL' = 90 - \angle C$ and $\angle L'RC = 90 - \angle B \implies L' = A'$ Claim: $R$ lie on $(AI_AI)$ By power of point $$DR.DA'=DB.DC=DI.DI_A$$ Which give us $(XLY) , (ABC)$ and $(AI_AI)$ intersect on $R$ Note if $(XLY)$ and $(AI_AI)$ are tangent to each other then $\angle LRA'$ is sum of angle made by respective circle tangents which is $\angle RYL + \angle RI_AA'$ $$\angle RYL = \angle RXB = 180 - \angle RDB$$ $$\angle RI_AA' = \angle IA'R + \angle IRA' = \angle IA'R + \angle II_AA' = \angle IA'R + \angle I_AIA' = \angle A'DI_A = \angle ADR$$ So $\angle RYL + \angle RI_AA' = 180 - \angle C - \frac{\angle A}{2}$ As we know $\angle LRA' = 180 - \angle C - \frac{\angle A}{2}$ We get $(XLY)$ and $(AI_AI)$ are tangent to each other.

Let $(LYZ)\cap (ABC)=T,AI\cap BC=X$ and the perpendicular line from $A$ to $BC$ intersect $(ABC)$ at $H$. Lemma: $ABC$ is a triangle with incenter $I$ and $AI$ intersects $BC,(ABC)$ at $X,M$ respectively. $A'$ is the antipode of $A$ on $(ABC)$. Also $A'X$ intersects $(ABC)$ at $E$ and $EI$ meets $(ABC)$ at $F$. If $N$ is the midpoint of arc $BAC$, then show that $NF\parallel A'I$. Proof: Let $MF\cap BC=R,A'I\cap (ABC)=S,AS\cap BC=P,PM\cap (ABC)=U$. $K$ lies on $BC$ such that $AK\perp BC$. Note that $S$ is $A-$sharky devil point and $\measuredangle PIA=90$. $U$ is $A-$mixtilinear touch point hence $N,I,U$ are collinear. Also we observe that under the inversion centered at $M$ with radius $MI$, $EIF$ swaps with $(MKIR)$. \[\measuredangle SMN=\measuredangle SUN=\measuredangle SUI=\measuredangle SPI=\measuredangle API=\measuredangle AKI=90-\measuredangle IKR=90-\measuredangle IMR=\measuredangle FMA'\]Thus, $NF\parallel SA'$.$\square$ Claim: $T,X,A'$ are collinear. Proof: \[\measuredangle LZY=\measuredangle CAI-\measuredangle ACL=\frac{\measuredangle A}{2}-\frac{\measuredangle B-\measuredangle C}{2}=90-\measuredangle B=\measuredangle HCB\]\[\measuredangle ZYL=\frac{\measuredangle A}{2}+\frac{\measuredangle B-\measuredangle C}{2}=90-\measuredangle C=\measuredangle CBH\]Hence $HBC\sim LYZ$. Also since $HD\perp BC$ and $LA\perp YZ$, we get $HBDC\sim LYAZ$. $T$ is the center of spiral homothety sending $BC$ to $YZ$ thus, $TBDCH\sim TYAZL$. \[\measuredangle XAT=\measuredangle YAT=\measuredangle BDT=180-\measuredangle TDX\]Which implies $A,X,D,T$ are concyclic. Note that $AX$ is diameter on this circle subsequently, $\measuredangle ATX=90=\measuredangle ATA'$. So $T,X,A'$ are collinear.$\square$ Claim: $A',I_A,I,T$ are concyclic. Proof: \[XI_A.XI=XB.XC=XA'.XT\]Which gives the expected result.$\square$ Claim: $(LYZT)$ and $(I_AA'IT)$ are tangent to each other. Proof: Let $AH\perp BC$ and $H\in (ABC),TI\cap (ABC)=W$. \[\measuredangle LZT+\measuredangle TA'I=(90-\measuredangle B+\measuredangle YZT)+\measuredangle TA'S=(90-\measuredangle B+\measuredangle BLT)+\measuredangle TLS=\measuredangle HLT+\measuredangle TLS=\measuredangle HLS\]By Lemma, we have that $LW\parallel A'S$. Apply Pascal on $LLWHA'S$ to get that $BC_{\infty},A'S_{\infty},WH\cap SL$ are collinear. Since this line must be at infinity, $SL\parallel HW$. Hence $\measuredangle LZT=\measuredangle HLS=\measuredangle LTW$ which gives the tangency of $(LYZT)$ and $(I_AITA')$ as desired.$\blacksquare$

200th post Solved with math_holmes 15 Really nice problem , though only angle chasing was involved, it was fun. $\rule{35cm}{0.1pt}$ Let $AM\cap BC=D$, $A'D\cap \odot(ABC)=R$ where $M$ is the $L-$antipode. Claim: $\odot(RYBD)$ is cyclic. \[\angle BRD=\angle BLA'=\angle BYD\]Claim: $\odot(RYLZ)$ is cyclic. \[\angle ZLR=\angle RBC=\angle ZYR\]Claim: $\odot(RIA'I_A)$ is cyclic. \[RD\cdot RA'=BD\cdot CD=DI\cdot DI_A\]Let $N$ be the midpoint of $AD$, then $\rule{35cm}{0.1pt}$ Claim: $NR$ is tangent to $\odot(RYLZ)$ and $\odot(RIA'I_A)$ and $NR=ND$ \begin{align*} \angle A'II_A=\angle A'RI_A=\angle A'RI=\angle A'I_AD=x\\ \angle A'I_AR=\angle A'DI_A=\angle RDN=\angle IRD=y \end{align*}then we have $\angle RIN=x+y$ and $NRD=y$ so $\angle NRI=y-x$ and similarly we also have $y-x=\angle A'I_AI$. So $NR$ is tangent to $\odot(A'I_ARI)$. Now we show $NR$ is also tangent to $\odot(LYZR)$. \begin{align*} \angle YRN &= \angle YRD - \angle NRD\\ &= \angle LBD - (\angle NRI + \angle IRA')\\ &=\left(90^{\circ}-\frac{\angle A}{2}\right)-(\angle IA'R + \angle IRA')\\&=\left(90^{\circ}-\frac{\angle A}{2}\right)-\angle RI_AA'\\&=\left(90^{\circ}-\frac{\angle A}{2}\right)-\angle IDR\\&=\widehat{LB}-\widehat{BM}-\widehat{AR}\\&=\widehat{RB}\\&=\angle RAB\\&=\angle RLY \end{align*}Thus $\odot (LYZ)$ is tangent to $\odot (A'II_A)$ at $R$ and we are done. $\rule{35cm}{0.1pt}$ [asy][asy] 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 = -19.71762211856687, xmax = 11.335548360553895, ymin = -12.376226640019961, ymax = 13.078331916213367; /* image dimensions */ draw((-6,4)--(-8,-2)--(0,-2)--cycle, linewidth(0.6)); draw(arc((-2.9196369730490948,-9.048627177541055),1.0432196129603841,79.6764268057746,103.282525588539)--(-2.9196369730490948,-9.048627177541055)--cycle, linewidth(0.6)); 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draw((-6,4)--(-2.9196369730490948,-9.048627177541055), linewidth(0.6)); draw((-4,4.472135954999577)--(-8,-2), linewidth(0.6)); draw((0,-2)--(-7.416407864998738,10), linewidth(0.6)); draw((-7.416407864998738,10)--(-6,4), linewidth(0.6)); draw(circle((-8.472135954999578,5.527864045000421), 4.595058410947224), linewidth(0.6)); draw(circle((-8.236067977499792,-5.4721359549995805), 6.407474392457677), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-2,-4), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-5.52786404500042,2), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-4,4.472135954999577), linewidth(0.6)); draw((-2,-4)--(-2.9196369730490948,-9.048627177541055), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-5.080363026950906,0.10435526754189517), linewidth(0.6)); draw((-5.291796067500631,1)--(-8.373545369999913,0.9338634249997817), linewidth(0.6)); draw((-4,4.472135954999577)--(-4,-4.472135954999577), linewidth(0.6)); draw((-8.373545369999913,0.9338634249997817)--(-2.9196369730490948,-9.048627177541055), linewidth(0.6)); draw(circle((-5.291796067500631,1), 3.0824588902380468), linewidth(0.6)); draw((-5.080363026950906,0.10435526754189517)--(-2,-4), linewidth(0.6)); /* dots and labels */ dot((-6,4),linewidth(3pt) + dotstyle); label("$A$", (-6.712150943660748,4.2109652060501155), NE * labelscalefactor); dot((-8,-2),linewidth(3pt) + dotstyle); label("$B$", (-8.659494221186797,-2.5351882910936903), NE * labelscalefactor); dot((0,-2),linewidth(3pt) + dotstyle); label("$C$", (0.31219445027250453,-2.430866329797652), NE * labelscalefactor); dot((-5.080363026950906,0.10435526754189517),linewidth(3pt) + dotstyle); label("$I$", (-4.938677601628095,0.3162786509980215), NE * labelscalefactor); dot((-2.9196369730490948,-9.048627177541055),linewidth(3pt) + dotstyle); label("$I_A$", (-2.5392724918192116,-9.281341788237496), NE * labelscalefactor); dot((-2,-4),linewidth(3pt) + dotstyle); label("$A'$", (-1.739470788549584,-4.552079542817096), NE * labelscalefactor); dot((-4,4.472135954999577),linewidth(3pt) + dotstyle); 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