Rewrite the problem in terms of the excentral triangle: Quote: Let $\triangle ABC$ have orthocenter $H$ and orthic triangle $\triangle DEF$, and let $G$ be the foot from $H$ to $EF$. If $DG$ intersects $AB,AC$ at $X,Y$ respectively, show $(AXY)$ and $(DHG)$ are tangent. Let $Q$ be the $A$-Queue point and $T=EF \cap BC$. We show the desired tangency happens at $Q$. Note $DHGQT$ all lie on the circle with diameter $TH$. This implies $Q$ is spiral center sending $BCD \to FEG$, and hence by gliding principle it's also the center of the spiral similarities $CD \to AX$ and $BD \to AY$. Now $$\measuredangle QXA = \measuredangle QDC = \measuredangle QDB = \measuredangle QYA$$says $QAXY$ cyclic, and finally$$\measuredangle QDT = \measuredangle QDB = \measuredangle QYA$$says that $(DHGQT)$ and $(QAXY)$ are in fact tangent at $Q$.

WLOG $BA<CA$. Let $S$ denote the miquel point of $BEFC$. We claim that $S$ is the desired tangency point. Note that by definition $S$ lies on $\odot (BIC)$, $\odot (BDE)$, $\odot (CDF)$ and $\odot (I_AEF)$ First we show that $S \in \odot (AID)$ Proof : $$\angle ADS = \pi - \angle SDF = \pi - \angle SCF = \pi - \angle SCI_A = \pi - \angle SII_A = \angle AIS$$ Now to prove that $S$ is the desired tangency point, we need to prove that $\angle SAI + \angle SFI_A = \angle ISI_A = \frac {\pi}2$. We have : $$ \angle SAI + \angle SFI_A = \pi - \angle IDS + \angle CFS =\pi - \angle IDS + \angle = \frac {\pi}2$$ We are done. $\square$

Let $M$ be the Miquel Point of quadrilateral $CDEI_A$. Claim 1. $AIDM$ is cyclic

Claim 2. $(AID)$ and $(EFI_A)$ are tangent at $M$. proof. This is equivalent to proving $\angle DME = \angle DAM + \angle EFM$.

By Claim 2, we are done.

Some solutions are bad, and some solutions are so bad. First, invert about the incircle. Then, invert about the image of $A$ with power $AI\cdot AI_A$ (normally $AD\cdot AH$). It is sufficient to solve the following double-inverted problem (which has been relabeled): Let $ABC$ be a triangle, and let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$, respectively. Let $H$ be the orthocenter. Let $M_A$ be the midpoint of $BC$, and let $M'$ be the inverse of $M_A$ in the (second) inversion circle, i.e. the $A$-Humpty point. Let $T$ be the midpoint of $AM'$. Suppose that line $DT$ intersects $(BDF)$ and $(CDE)$ at $P$ and $Q$, respectively. Show that $(HPQ)$ is tangent to $AM_A$ (at $M'$). I claim that $HM'$ is the diameter of $(HPQ)$, which solves the problem since $\angle AM'H=\angle ADM_A=90^\circ$. Since $BH$ is the diameter of $(BDHF)$, $\angle BDH=90^\circ$, so we just need $B$, $P$, $M'$ collinear ($C$, $Q$, $M'$ collinear is analogous.). Notice that $T$ is the midpoint of the $A$-symmedian chord in $\triangle AFE$, so $T$ is the Dumpty point of $\triangle AFE$, and we have that $T$ lies on the "$BOC$" circle of $\triangle AEF$, which is the nine-point circle of $\triangle ABC$. Also, by Three Tangents Lemma, $DM_B$ is tangent to $(BDF)$, where $M_B$ is the midpoint of $AC$. Finally, we have (in ELSMO style) \[\angle PBD\stackrel{(1)}{=}\angle PDM_B\stackrel{(2)}{=}\angle TDM_B\stackrel{(3)}{=}\angle TM_AM_B\stackrel{(4)}{=}\angle TAB\stackrel{(5)}{=}\angle M_AAB\stackrel{(6)}{=}\angle B'BC\stackrel{(7)}{=}\angle M'BD,\]as desired. Footnotes: (1). Tangent (2). Def. (Collinear) (3). Nine-point circle gives cyclic quad (4). $M_BM_A\parallel AB$ (5). Def. (Collinear) (6). Humpty Points property (7). Def. (Collinear)

Let $M$ be a Miquel Point of $CDEI_{A}$ and $MI_{A}\cap BC=G$. $$\measuredangle AIM=\measuredangle I_{A}IM=\measuredangle I_{A}BM=\measuredangle EDM=\measuredangle ADM\implies M\in \odot (AID)$$$$\measuredangle IMG=\measuredangle IMI_{A}=90^{\circ}=\measuredangle IDG\implies G\in \odot (AID)$$Finally $\measuredangle EMD=\measuredangle EBD=\measuredangle MGD+\measuredangle EI_{A}M$, so $M$ is desired tangency point.

[asy][asy] size(10cm); 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 = -17.875625000000007, xmax = 19.75562500000001, ymin = -14.19312499999996, ymax = 7.913124999999983; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-1.2661247310229125,-5.705843708998829), 3.9623288136575754), linewidth(0.8) + zzttff); draw(circle((-9.417880604714508,-2.884368109939817), 6.6376989476256325), linewidth(0.8) + linetype("4 4") + blue); draw(circle((-2.653087785208155,-9.71303208493889), 2.9744326356546784), linewidth(0.8) + zzttff); draw((-4.29125,1.3318750000000001)--(-4.74,-3.8), linewidth(0.8) + qqwuqq); draw((-4.74,-3.8)--(2.24,-3.86), linewidth(0.8) + qqwuqq); draw((-2.8308770831025636,-2.065569034719119)--(-2.8459261868452876,-3.8162814367892963), linewidth(0.8)); draw((-4.29125,1.3318750000000001)--(0.29862762105674,-9.346118383278538), linewidth(0.8) + ffvvqq); draw((-4.74,-3.8)--(0.29862762105674,-9.346118383278538), linewidth(0.8)); draw((-0.5891337021448895,-11.854840444025694)--(-4.29125,1.3318750000000001), linewidth(0.8) + ffvvqq); draw((-4.746423728588502,-7.599931488138171)--(-2.428377034365336,-5.303566388058189), linewidth(0.8)); draw((-4.29125,1.3318750000000001)--(2.24,-3.86), linewidth(0.8) + qqwuqq); draw(circle((-3.8092387715904703,-5.701544611918473), 2.117117940839996), linewidth(0.8) + linetype("4 4") + zzttff); draw((2.24,-3.86)--(-0.5891337021448895,-11.854840444025694), linewidth(0.8)); /* dots and labels */ dot((-4.29125,1.3318750000000001),dotstyle); label("$A$", (-4.1225,1.75375), NE * labelscalefactor); dot((-4.74,-3.8),dotstyle); label("$B$", (-4.586562500000002,-3.3931249999999875), NE * labelscalefactor); dot((2.24,-3.86),dotstyle); label("$C$", (2.416562500000001,-3.435312499999988), NE * labelscalefactor); dot((-2.8308770831025636,-2.065569034719119),linewidth(4pt) + dotstyle); label("$I$", (-2.6459375000000014,-1.747812499999992), NE * labelscalefactor); dot((0.29862762105674,-9.346118383278538),linewidth(4pt) + dotstyle); label("$I_A$", (0.47593750000000007,-9.004062499999973), NE * labelscalefactor); dot((-2.8459261868452876,-3.8162814367892963),linewidth(4pt) + dotstyle); label("$D$", (-2.688125000000001,-3.4775), NE * labelscalefactor); dot((-2.0054634294322087,-6.809959196109981),linewidth(4pt) + dotstyle); label("$E$", (-1.8443750000000008,-6.472812499999979), NE * labelscalefactor); dot((-0.5891337021448895,-11.854840444025694),linewidth(4pt) + dotstyle); label("$F$", (-0.41,-11.535312499999966), NE * labelscalefactor); dot((-4.746423728588502,-7.599931488138171),linewidth(4pt) + dotstyle); label("$X$", (-4.586562500000002,-7.274374999999978), NE * labelscalefactor); dot((-2.428377034365336,-5.303566388058189),linewidth(4pt) + dotstyle); label("$K$", (-2.26625,-4.954062499999984), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $X$ be the center of spiral sim. sending $\overline{BE}$ to $\overline{CF}$. Then by Miquel point, $X$ is the intersection of the circles $(BDE)$, $(BCII_A)$, $(CDF)$,$(EI_AF)$. Hence $$\angle ADX=180^{\circ}-\angle XDE=180^{\circ}-\angle XBI_A=180^{\circ}-\angle XII_A=\angle AIX$$Hence $X$ lies on $(ADI)$ as well. \newline\newline Now let the tangent of $(ADI)$ at $X$ intersects $EF$ at $K$. Notice that $$\angle EXA=\angle AXD+\angle EXD=\angle DII_A+\angle CBI_A=90^{\circ}-\frac{C}{2}=\angle DCI_A=180^{\circ}-\angle DXF$$Therefore, $$\angle KXF=\angle DXF-\angle DXK=180^{\circ}-\angle EXA-\angle XAD=\angle KEX$$which implies $KX$ is tangent to $(EI_AF)$ as well.

Apparently this problem was generated by an AI. Let \(I_B\) and \(I_C\) be the \(B-\) and \(C\)-excenters, and let \(S=\overline{BC}\cap\overline{I_BI_C}\). Let \((I_AI_BI_C)\) and \((BICI_A)\) intersect again at \(L\). By radical axis theorem \(I_A\), \(L\), \(S\) collinear, and also \(\angle IAS=\angle IDS=\angle ILS=90^\circ\), so \(A\), \(I\), \(D\), \(L\), \(S\) are concyclic. We claim \(L\) is the desired tangency point. Let lines \(BI\) and \(CI\) meet \((AID)\) again at \(E'\) and \(F'\). Since \(AIBI_C\) is cyclic, we have \(\overline{SE'}\parallel\overline{BI_C}\) by Reim's. Analogously \(\overline{SF'}\parallel\overline{CI_B}\). [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen tri=heavygreen; pen qua=lightred; pen fil=invisible; pen fil2=invisible; pen tfil=invisible; pen qfil=invisible; pair A,B,C,I,IA,IA,IB,IC,D,EE,F,SS,L,Ep,Fp; A=dir(140); B=dir(210); C=dir(330); I=incenter(A,B,C); IA=2dir(270)-I; IB=extension(B,I,C,IA); IC=extension(C,I,B,IA); D=foot(I,B,C); EE=extension(B,IA,A,D); F=extension(C,IA,A,D); SS=extension(B,C,IB,IC); L=foot(I,IA,SS); Ep=extension(B,I,L,EE); Fp=extension(C,I,L,F); draw(IA--SS,orange); draw(EE--Ep,heavycyan+dashed); draw(F--Fp,heavycyan+dashed); draw(I--Ep,lightblue); draw(C--Fp,lightblue); draw(IC--SS,qua); filldraw(circumcircle(IA,IB,IC),qfil,qua+dashed); filldraw(IA--IB--IC--cycle,qfil,qua); draw(F--IA,qua); filldraw(circumcircle(L,EE,F),tfil,tri+dashed); filldraw(circumcircle(A,I,D),tfil,tri); draw(A--F,tri); filldraw(circumcircle(B,I,C),fil2,pri2); draw(IA--SS,fil); filldraw(A--B--C--cycle,fil,pri); draw(B--SS,pri); dot("\(A\)",A,dir(80)); dot("\(B\)",B,dir(215)); dot("\(C\)",C,E); dot("\(I\)",I,dir(300)); dot("\(D\)",D,SE); dot("\(I_A\)",IA,SE); dot("\(E\)",EE,NE); dot("\(F\)",F,SE); dot("\(I_B\)",IB,NE); dot("\(I_C\)",IC,N); dot("\(L\)",L,dir(260)); dot("\(S\)",SS,W); dot("\(E'\)",Ep,S); dot("\(F'\)",Fp,NW); [/asy][/asy] Finally, \begin{align*} \measuredangle I_ALE&=\measuredangle I_AFE=\measuredangle CI_BI_C+\measuredangle SAD\\ &=\measuredangle DIB+\measuredangle SID=\measuredangle SIE'=\measuredangle SLE', \end{align*}so \(L\in\overline{EE'}\), and similarly \(L\in\overline{FF'}\). A homothety at \(L\) sends \((AID)\) to \((I_AEF)\), so they are tangent at \(L\).

Let $M$ be the miquel point of $DEI_aC$. Just angle chase to prove $AIDM$ cyclic and then we claim $M$ is the point of contact of the two circles. Then, taking the tangent to $AIDM$ at $M$, we'll prove that it is also tangent to $MEI_aF$. Really little angle chase (see 2) to finish the tangency. 2:

We claim that the tangency point is $A'I \cap (BIC)$ where $A'$ is the midpoint arc of $BC$ containing $A$. Let $X = A' I \cap (BIC)$. Claim 01. $AIDX$ is cyclic. Proof. Let $M$ be the antipode of $A'$. By Incenter Excenter Lemma, $M$ is the center of $(BIC)$. Thus, \[ \measuredangle A'XI_A = \measuredangle IXI_A = 90^{\circ} = \measuredangle A'AM = \measuredangle A'AI_A \]since $A'M$ is the diameter of $(ABC)$ by definition. Hence, $A'AXI_A$ is cyclic. By Radical Axis Theorem on $(AA'BC), (BCXI_A), (AA'XI_A)$, we have $AA', BC, XI_A$ concur. Suppose they concur at $Y$. It is immediate to see that $YAIX$ is cyclic since $\measuredangle IAY = 90^{\circ} = \measuredangle IXY$. However, $\measuredangle IDY = 90^{\circ} = \measuredangle IAY$ as well. Therefore, $AIDXY$ is cyclic. Claim 02. $XBDE$ and $XCDF$ are both cyclic. Proof. Note that \[ \measuredangle XDE = \measuredangle XDF = \measuredangle XDA = \measuredangle XIA = \measuredangle XII_A = \measuredangle XBI_A = \measuredangle XBE. \] Claim 03. $XEI_AF$ is cyclic. Proof. Note that \[ \measuredangle XI_A E = \measuredangle XI_A B = \measuredangle XCB = \measuredangle XCD = \measuredangle XFD = \measuredangle XFE \] Therefore, we have $X \in (AID) \cap (EI_AF)$ from our previous claims. Now, draw a line $\ell$ tangent to $(AIDX)$ at $X$. We'll prove that $\ell$ tangent $(XEI_A)$ as well. It suffices to prove that $\measuredangle YAX = \measuredangle I_AFX$, but this is true because \[ \measuredangle YAX = \measuredangle YDX = \measuredangle BDX = \measuredangle BEX = \measuredangle I_A EX = \measuredangle I_A FX \]Hence, $(AID)$ and $(EI_AF)$ are tangent at $X$.

PROA200 wrote: Some solutions are bad, and some solutions are so bad. and some solutions are even worse. Let $L$ be the midpoint of arc $BAC$, $T = \overline{LI} \cap (ABC)$, and $I'$ be the reflection of $I$ over $T$. I claim that $I'$ is the desired point of tangency. Let $M_A$ be the center of $(BIC)$. Since $\overline{M_AT} \perp \overline{IT}$, we have $M_A=M_AI'$ so $I'$ lies on $(BIC)$. Because $L$ is the intersection of tangents at $B$ and $C$ to $(BIC)$ and $I$ is the antipode of $I_A$, $I'$ is actually the reflection of the $I_A$-Queue point over the perpendicular bisector of $\overline{BC}$. Lemma: Let $M$ be the midpoint of $\overline{BC}$, then $\overline{I_AM} \parallel \overline{AD}$. Proof. Note that $\overline{DA}$ is the $D$-symmedian in the intouch triangle, and since $\overline{ABC}$ is the orthic triangle of the excentral triangle, $\overline{I_AM}$ is the $I_A$-symmedian in the excentral triangle. Since the intouch triangle and the excentral triangla are homothetic, the conclusion follows. $\square$ Lemma: $AIDI'$ is cyclic. Proof. For ratio lemma reasons, lines $\overline{I'D}$ and $\overline{I_AM}$ meet at a point on $(BIC)$. Then $\angle II'D = \angle II_AM = \angle IAD$ as desired. $\square$ Now rephrasing wrt $\triangle I_ACB$, we have the following problem (points relabeled): "In acute triangle $ABC$ with $AB<AC$, let $M$ be the midpoint of $\overline{BC}$, $Q$ be the $A$-Queue point, and $Q'$ be the reflection of $Q$ over the perpendicular bisector of $\overline{BC}$. Let $A'$ be the $A$-antipode and $P$ be the projection of $A'$ on $\overline{BC}$. The line through $P$ parallel to $\overline{AM}$ meets $\overline{AC}$, $\overline{AB}$ at $E$, $F$ respectively. Prove that the $(AEF)$ and $(Q'PA')$ are tangent at $Q'$." Let $X$ be the foot from $A$ to $\overline{BC}$, which is also the reflection of $P$ over $M$. Let $Y = \overline{AX} \cap \overline{Q'M}$ and $J = \overline{AM} \cap \overline{Q'P}$, both of which lies on $(ABC)$ by ratio lemma. Let $K$ be the point such that $ABKC$ is harmonic. Claim: $Q'$ is the center of the spiral sim sending $\triangle BKC$ to $\triangle FAE$. This means it lies on $(AEF)$ by Miquel points. Proof. Since $\angle AEF = \angle CAM = \angle KAB = \angle KCB$ and similarly $\angle AFE = \angle KBC$, triangles $BKC$ and $FAE$ are indeed similar. I will prove that $Q'A/Q'K$ is the similarity ratio, which solves the problem because obviously $Q'$ lies on the same side of $\overline{AK}$ as $C$, and there is only one such point on $(ABC)$. (this spiral center must lie on $(ABC)$ by Miquel points) I will first show that $\triangle PMJ \sim \triangle AQ'K$, which is just angle chasing: it follows from $\angle AKQ' = \angle AJQ' = \angle MJP$ and $\angle PMJ = \angle CAJ + \angle BCA = \angle KAB + \angle BCA = \angle KQ'A$. Now, we find $$\frac{EF}{CB} = \frac{d(A,\overline{EF})}{d(K,\overline{CB})} = \frac{d(M,\overline{EF})}{d(J,\overline{CB})} = \frac{d(X,\overline{AM})}{d(J,\overline{CB})} = \frac{XM}{JM} = \frac{PM}{JM} = \frac{AQ'}{KQ'}$$as desired. $\square$ Let $O$ be the center of $(ABC)$, $R$ be the center of $(A'PQ')$ and $S$ be the center of $(AEQ'F)$, so that it suffices to prove that $S$, $Q'$, $R$ are collinear. This is just angle chasing: \begin{align*}\angle SQ'R &= \angle SQ'O + \angle OQ'A' + \angle A'Q'R \\ &= \angle AQ'K + \angle AA'Q' + (\angle Q'PA' - 90^\circ) \\ &= \angle Q'CK + \angle CPQ' \\ &= \angle Q'CK + \angle CBQ' + \angle BAJ \\ &= \angle Q'CK + \angle CBQ' + \angle KAC \\ &= 180^\circ. \end{align*}$\blacksquare$

Denote circumcircle of $\triangle AID,\triangle BIC,\triangle EFI_A$ by $\Gamma,\Omega,\gamma$ respectively. Define $S$ by an antipode of $I$ on $\Gamma$. Let a line $AI$ intersect $\gamma$ again at $J$. By angle chasing, $\measuredangle IBC=\measuredangle II_AC=\measuredangle JEF$ and $\measuredangle ICB=\measuredangle II_AB=\measuredangle JFE$, so $\triangle JEF\stackrel{+}{\sim}\triangle IBC$. Because $\measuredangle JAE=\measuredangle IAD=\measuredangle ISB$, $\triangle JEF\cup A\stackrel{+}{\sim}\triangle IBC\cup S$. Next, let $IS$ intersect $\Omega$ again at $U$. Then, $\triangle JEF\cup A\cup I_A\stackrel{+}{\sim}\triangle IBC\cup S\cup U$. Consider the spiral similarity that send the former system to the latter one. By well-known lemma, the center of that spiral similarity is the second intersection of $\odot(SIA)=\Gamma$ and $\odot (IUI_A)=\Omega$, called this intersection $T$. Since $T$ lies on $\odot(IBC)$, it must also lie on $\odot(JEF)=\gamma$. By angle chasing $\measuredangle STI=90^{\circ}=\measuredangle I_ATI$, thus $S,T,I_A$ are collinear. Moreover by the similarity, $\measuredangle SAT=\measuredangle IJT=\measuredangle I_AFT$. Hence, $\overset{\Large\frown}{ST}$ and $\overset{\Large\frown}{TI_A}$ are bases of the same angle, which implies that $\Gamma$ and $\gamma$ are tangent as desired.

Here's my solution. Let $N$ be the midpoint of the arc $BAC$. We claim that the desired tangency point is the refletion of the incenter $I$ over $T\equiv NI\cap (ABC)$, i.e. $NI\cap (BIC)$. Let this point be $K$. $\textbf{Lemma 1.\hspace{2 mm}}$ $K\in (AID)$. $\textbf{Proof.\hspace{1 mm}}$ We use complex numbers with the assumption that $(ABC)$ is the unit circle . Let $A, B, C, I, N$ have complex coordinates $a^2, b^2, c^2, -(ab+bc+ca), bc$ respectively. Since $D$ is the projection of $I$ onto $BC$ we have $$d=\frac{a(b^2+c^2)+bc(b+c)-a^2(b+c)}{2a}$$From $I=n+t-nt\bar{I}$ we have $$t=\frac{-a(ab+ac+2bc)}{2a+b+c}$$By the definition of $K$ $$k-(ab+bc+ca)=\frac{-2a(ab+ac+2bc)}{2a+b+c}\to k=\frac{a(b^2+c^2)+bc(b+c)}{2a+b+c}$$It remains to check $$\frac{(k-d)(I-a^2)}{(k-a^2)(I-d)}\in \mathbf R\hspace{5 mm} (*)$$Luckily, these differences have nice forms $$k-d=\frac{(a+b)(a+c)(b+c)(2a-b-c)}{2a(2a+b+c)}$$$$k-a^2=-\frac{(a+b)(a+c)(2a-b-c)}{2a+b+c}$$$$I-a^2=-(a+b)(a+c)$$$$I-d=-\frac{(a+b)(a+c)(b+c)}{2a}$$Hence $(*)$ is $$-\frac{b+c}{2a}\cdot \frac{2a}{b+c}=-1 \hspace{3 mm} \textsf{which is real, done.}\hspace{2 mm} \Box$$ $\textbf{Lemma 2.\hspace{2 mm}}$ $K\in (I_{A}EF)$. $\textbf{Proof.\hspace{1 mm}}$ It's enough to show that the quadrilateral $KDCF$ is cyclic. But by $\textsf{lemma 1}$ $$\angle KDF=\angle KII_{A}=\angle KCF$$done. $\Box$ $\textbf{Lemma 3.\hspace{2 mm}}$ $(AIDK)$ and $(I_{A}EFK)$ are tangent. $\textbf{Proof.\hspace{1 mm}}$ Using $\textsf{lemma 2}$ observe easily that $BKED$ is cyclic. $$\angle DKE=\angle CBI_{A}=90-{\angle B}/2=\angle KID+\angle KFE$$Consequently the desired tangency holds. $\mathcal{Q.E.D.}$

Posting for storage Define $X'$ as the Miquel point of $BDI_AF$ Part-1-$X'\in (ADE)$ Proof $$\angle ADX = 180 - \angle XDE = 180 - \angle XBE = 180 - \angle XII_A = \angle AIX$$ Part-2-It suffice to show $$\angle DXE = \angle DAX + \angle EFX$$Proof $$\angle DXE = 90-\frac{B}{2}$$$$\angle XAD=\angle XID=\angle XIC-\angle DIC=\angle XBD-(90-\frac{C}{2})=\angle XI_AF-(90-\frac{C}{2}) \; $$$$\angle EFX=\angle EI_AX \implies (\angle XI_AF+\angle EFX)-90+\frac{C}{2}=90+\frac{A}{2}-90+\frac{C}{2}=90-\frac{B}{2}$$

Clean inversion! Let $G$ be the intersection of the $A$-external angle bisector with $BC$ and $T$ be the reflection of $I$ over the $A$-mixtilinear touchpoint of $\triangle ABC.$ It's well-known that $TGDAI$ is cyclic, $G,T,I_A$ are collinear, and $TBIC$ is harmonic. Observe that $\measuredangle TDF = \measuredangle TII_A = \measuredangle TCF,$ so $TDCF$ is cyclic and thus $T=(DCF)\cap (I_ACB)$ is the Miquel Point of $DEI_AC.$

TheUltimate123 wrote: Apparently this problem was generated by an AI. Is this true or just a joke?

Mathsisnice wrote: TheUltimate123 wrote: Apparently this problem was generated by an AI. Is this true or just a joke? Read the title of this post; it's the truth

Let the $A$-external angle bisector meet $\overline{BC}$ at $X$. Now let $T = \overline{XI_A} \cap (BIC)$. We claim that $T$ is the tangency point of the two given circles. It is easy to see that $A$, $D$ and $T$ all lie on a circle with diameter $XI$. Claim: $T$ lies on $(I_AEF)$. Proof. Let $I_B$ and $I_C$ be the $B$ and $C$ excenters. Since $I_B$, $I_C$, $B$ and $C$ are concyclic, by Miquel's theorem in $\triangle XI_CI_A$, we see that $XI_BBT$ is also cyclic. By angle chasing, one has \[ \measuredangle BTD = \measuredangle BTX + \measuredangle XTD = \measuredangle BI_BA + \measuredangle XAD = \measuredangle BED, \]and hence $B$, $T$, $E$, $D$ are concyclic. Similarly, $C$, $T$, $F$, $D$ are concyclic, so by Miquel's theorem again in $\triangle CDF$, it follows that $X$ lies on $(I_AEF)$. $\square$ Now the required tangency easily follows as \[ \measuredangle DXT + \measuredangle TI_AE = \measuredangle DBE = \measuredangle DTE. \]

TheUltimate123 wrote: Apparently this problem was generated by an AI. Where can we find more information about this?

zuss77 wrote: TheUltimate123 wrote: Apparently this problem was generated by an AI. Where can we find more information about this? This exists, but I'm not sure if it's the exact AI used to create this problem. EDIT: ok it is the exact AI

Let $\omega$ denote the circumcircle of $\triangle AID$. We will define some of our auxiliary points. Let $D'$ be the second intersection of $\overline{BC}$ with $\omega$, $E'$ be the intersection of $\overline{CI}$ and $\omega$, and $F'$ be the intersection of $\overline{BI}$ and $\omega$. We realize that $\overline{D'E'} \parallel \overline{I_CBI_A}$ because they're both perpendicular to $\overline{IB}$. Now, let $X$ be the foot of the perpendicular from $I$ to $\overline{D'I_A}$ Let $E'' = \overline{E'X} \cap \overline{I_AB}$ and $F'' = \overline{F'X} \cap \overline{I_AC}$; $XEI_AF$ is a cyclic quadrilateral that is similar to $XE'D'F'$. It is evident (by homothety with center $X$) that the circumcircles of these two quadrilaterals are tangent. We would now like to show that $E = E''$ and $F=F''$. The perpendicular condition gives us that $X$ is on $\omega$ and the circumcircle of $\triangle IBC$. We will now show the following three cyclic quadrilaterals with angle chases: Claim: $BDXE''$ is cyclic. Proof: $\angle D'DX = 180 - \angle D'E'X = 180 - \angle XE''I_A = \angle XE''B$. Claim: $BXI_CD'$ is cyclic. Proof: $\angle BXI_A = 180 - \angle BCI_A = 180 - \angle I_AI_CI_B = \angle BI_CD'$. Claim: $E''XI_CA$ is cyclic. Proof: $$\angle E''XI = 90 - \angle E''XI_A = 90 - \angle D'XE' = 90 - \angle D'DI' = \angle E'D'I$$$$\angle IXA = \angle AD'I$$$$\angle E''XA = \angle E'D'A = \angle BI_CA \because \overline{BI_C} \parallel \overline{E'D'}$$ We now basically have a complete quadrilateral, and we can use angle chasing to show that $A,D,E''$ are definitively collinear (by analogous reasoning, $A,D,F''$ would also be collinear) $$\angle ADB = \angle ADD'$$$$\angle BDX = 180 - \angle XE'D'$$$$\angle XDE'' = \angle XBE'' = \angle I_CD'X = \angle AD'I_A = \angle AE'X$$$$\angle BDE'' = 180 + \angle XE'A - \angle XE'D' = 180 - \angle AE'D' = 180 - \angle ADD'$$ So, this shows that $\angle ADB + \angle BDE'' = 180$, which proves that $E'' = E$. $\blacksquare$.

Let $G$ be the miquel point of $CDEI_A$. The claim is that $G$ is the desired tangency point. Claim: $AIDG$ is cyclic. Proof. We angle chase. Notice that \begin{align*} &\measuredangle DGI=\measuredangle DGB-\measuredangle IGB=\measuredangle DEB-\measuredangle II_AB \\ &=\measuredangle AEI_A-\measuredangle AI_AE=\measuredangle EAI_A=\measuredangle DAI \end{align*}proving the claim. $\blacksquare$ Claim: The circles $I_AEFG$ and $AIDG$ are tangent at $G$. Proof. First, notice that $90^{\circ}=\measuredangle BDG+\measuredangle GDI=\measuredangle BDG+\measuredangle GAI$ as $ID\perp BC$. Then, let $\ell$ be the tangent line to $(EI_AFG$ at $G$. We have that \[\measuredangle(\ell, GI)=90^{\circ}-\measuredangle(I_AG,\ell)=90^{\circ}-\measuredangle I_AEG=90^{\circ} -\measuredangle BEG =90 ^{\circ} -\measuredangle BDG=\measuredangle GAI\]finishing. $\blacksquare$ We are thus done.

WLOG $AB<AC$. Let $N$ be the midpoint of major arc $BAC$. Let $H$ be the orthocenter. Let $S=ND\cap (ABC)$. Then $\angle ASD = \angle ASN = \frac{\angle B - \angle C}{2}$. We can also compute $$180-\angle AID = \angle IAH = \frac{\angle A}{2}-(90-\angle B)=\frac{\angle B - \angle C}{2},$$Thus $AIDS$ are concyclic. Now let $X=NI\cap (AIDS) \neq I$. Now we have $\angle NBC = \angle NCB = \angle NSD$, so $NB$ is tangent to $(BSD)$, and $NB^2=NS\cdot ND = NI \cdot NX$. Since it is well-known that $NB$ is tangent to $BIC$, we have by Power of a Point that $X \in (BIC)$. Since $NB$ and $NC$ are tangent to $(BICX)$, and $N,I,X$ are collinear, we have $(B,C;I,X)=-1$.Now let $R$ be the midpoint of $NI_A$. Consider the circle $\omega$ centered at $R$ with diameter $NI_A$. $NA \perp AI_A \rightarrow A \in \omega$. Also, we have $90 = \angle IXI_A = \angle NXI_A$, so $X \in \omega$. Now let $G$ be the midpoint of $EF$. Here is a brief interlude showing $G \in \omega$. Here are coordinates of various points. They can be computed in this order. $D=(0:s-c:s-b)$ $I_A=(-a:b:c)$ $E=-a(s-c):b(s-c):b(s-b)$ $F=-a(s-b):c(s-c):c(s-b)$ $G=-2a^2:b^2-c^2+ab+ac:c^2-b^2+ab+ac$ Now using the formula for perpendicular bisectors, $a^2(z-y)+x(c^2-b^2)=0$ we can easily check $G$ is on the bisector of $BC$. We can similarly straightforwardly check that $GI_A || BC$ since in normalized barycentric coordinates they both have $A$ coordinate $\frac{-a}{-a+b+c}$. This implies $NG \perp GI_A$, so $G \in \omega$. Now projecting through $I_A$ we get $-1=(B,C;I,X)=(E,F;A,Y)$, where $Y=I_AX \cap AD$. We now have by Power of a Point that $XY \cdot YI_A = YG \cdot YA = YE \cdot YF$. This implies that $EFI_AX$ is cyclic. Now angle chasing gives $$\angle RAI = \angle RI_AA = \angle NI_AA = \angle NXA = \angle IXA,$$so $RA$ is tangent to $AIDX$. Since $RA=RX$, $RX$ is the other tangent. Now we do one last angle chase to show $RI_A$ is tangent to $EXFI_A$. Let $f(\ell)$ be the unit complex number corresponding to the direction of $\ell$. Equalities are modulo rotation by $\pi$. Now it suffices to check $$f(RI_A)f(EF)=f(I_AF)f(I_AE)$$$$\iff f(I_AB)f(I_AC)=f(RI_A)f(AG)$$$$\iff f(IB)f(IC)=f(GI_A)f(AN)$$$$\iff \frac{f(IB)}{f(AN)}=\frac{f(BC)}{f(IC)}.$$Both of these are equal to $\frac{\angle C}{2}$, so we are done. Now $RI_A$ is tangent to $EI_AFX$, which means $RX$ is as well. Since $RX$ is tangent to both circles, we are done.

The whole point of the problem is classifying $T$, the point of tangency. We set $T = \overline{NI} \cap (BIC)$ where $N$ is the major arc midpoint. Let $M$ be the other arc midpoint. Since $NB$ and $NC$ are tangent to $(BIC)$, we observe that $(B, C; I, T) = -1$. $\textbf{Claim.}$ $AIDT$ is cyclic. $\textit{Proof.}$ Let $\overline{I_aT} \cap \overline{BC} = S$. Perspectivity at $I_a$ reveals that $S$ is the intersection point of the external bisector of $BAC$ with $BC$. Thus, $SA \perp AI$, and since $ST \perp TI$ we have $SATI$ cyclic. But then since $ID \perp DS$ we also have $D$ on this circle too. $\square$ $\textbf{Claim.} TEFI_a$ is cyclic. $\textit{Proof.}$ Let $E' = (BTD) \cap \overline{BI_a}$. We will show $E = E'$, and by symmetry we get a similar result for $F$. Indeed this isn't hard to show: \[ \measuredangle TDE' = \measuredangle TBE' = \measuredangle TIM = \measuredangle TIA = \measuredangle TDA, \]so $E'$ lies on $AD$. Thus $E' = E$. It now follows that $T$ is the center of spiral similarity taking $BC$ to $EF$, whence $TEFI_a$ is cyclic. $\square$ To finish, since we have $S-T-I_a$ just simply note that \[ \measuredangle SDT = \measuredangle BDT = \measuredangle BET = \measuredangle I_aET \]which shows that the arcs have the same measure, hence the tangency is proven.

Let $L$ and $M_A$ be the midpoints of arcs $BC$ containing and not containing $A$, respectively. Suppose that $LI$ meets $(BICI_A)$ again at $X$. We claim $X$ is the desired tangency point. First, we show that $AIDX$ is cyclic. Note that $\angle LAI_A = \angle LXI_A = 90^{\circ}$, so $LAXI_A$ is cyclic. Then by the Radical Center theorem, we know that $LA, BC, XI_A$ concur at some point $Y$. Thus $A, D, X$ all lie on the circle with diameter $IY$, so $AIDX$ is cyclic. Next, we show that $I_AFXE$ is cyclic. Observe that $\angle BXD = \angle IXD + \angle IXB = \angle DAI + \frac{1}{2}\angle C$. But $\angle BEA = 90^{\circ} - \angle(BI, AD) = 90^{\circ} - (\angle(BI, AI) -\angle DAI)$, and the right-hand-side reduces to $90^{\circ} - (90^{\circ} -\frac{1}{2}\angle C -\angle DAI) = \angle DAI + \frac{1}{2}\angle{C}$. Thus $BDEX$ is cyclic, and so is $CDFX$, by symmetry, and $X$ must be the spiral center mapping $FE$ to $CB$, which implies $XEI_AF$ is cyclic. Finally, let $LX$ meet $(FI_AEX)$ again at $Z$. Showing that the circumcircles of $AID$ and $I_AEF$ are tangent is equivalent to showing $IY \parallel ZI_A$. But $\angle YIX = \angle YDX= \angle BEX = 180^{\circ} - \angle XEI_A = \angle XZI_A$, so we are done.

Here's another complex bash (first one was post #13) using a different unit circle and a different characterization of the tangency point, and much less synthetic work.

first, by miquel point, circles $BI_AC$, $BDE$, and $EI_AF$ intersect at one point $G$, and it is obvious $I$ lies on $BI_AC$ we can also easily prove that $II_A$ is a diameter of $BI_AC$, so $IG$ is perpendicular to $GI_A$ construct a point $J$ such that $JE$ is perpendicular to $I_AE$, and that $JF$ is perpendicular to $I_AF$, so $J$ lies on $EI_AF$ then, $JG$ is perpendicular to $GI_A$ and $JGI_A$ are collinear extend $GI_A$ to intersect $BC$ at $H$, so we get that $IGH=90$, and since $ID$ is perpendicular to $DH$, $IDGH$ are concyclic then, $HIG=HDG=BEG=GJI_A$, so $IH$ and $JI_A$ are parallel and homothety yields that the circumcircles of $\triangle AID$ and $\triangle I_AEF$ are tangent to each other

Let $T = (BDE) \cap (I_AEF)$. Then, by Miquel's theorem, we see that $(CDF)$ intersects $(BDE)$ at $T$ as well. Furthermore as \[\angle BTC = \angle BTD + \angle DTC = \angle BED + \angle DFC = \angle FEI_A + \angle EFI_A = \angle BI_AC\] $T$ lies on $(BIC)$ too. Moreover, as \[\angle ITD = \angle BTD - \angle BTI = \angle BEA - \angle BI_AI = \angle IAD\] $T$ lies on $(AID)$ too. Let $\ell$ be the tangent at $T$ to $(ATD).$ First we notice, \[\angle TAI = 180 - \angle IDT = 180 - (90 + \angle BDT) = 90 - \angle BDT = 90 - \angle TFC.\] To finish, $$\angle(\ell, TI_A) = \angle ITI_A - \angle IAT = 90 - \angle IAT = \angle TFC$$ implying $\ell$ is the tangent to $(I_AEF)$ as needed.

A similar one. Let $N$ be the midpoint of the arc $BAC$ on $(ABC)$. $NA\cap BC=K,KI_A\cap (BI_AC)=P,PI\cap (BI_AC)=S$ Since $KI\perp SI_A,I_AI\perp KA,IP\perp KI_A$ we get that $(K,N,I_A,I)$ is an orthogonal system. Hence $N,I,P$ are collinear. $K,P,D,I,A$ lye on the circle with diameter $KI$. \[\angle BPD=\angle BPI+\angle IPD=\frac{\angle C}{2}+\angle IAD=\angle EI_AI+\angle I_AAE=\angle BEA=\angle BED\]Thus, $P\in (BED)$. Since $P$ is on $(BCI_A)$ and $(BDE),$ it is the miquel point of $DEI_AC$. So $P\in (EFI_A)$. \[\angle PI_AE+\angle DKP=\angle DBE=\angle DPE\]If $l$ is tangent to $PEI_A$ at $P,$ then \[\angle DPE-\angle (PD,l)=\angle (l,PE)=\angle PI_AE=\angle DPE-\angle DKP\]Hence $l$ is also tangent to $(DKP)$. These show that $(KPDIA)$ and $(PEI_AF)$ are tangent to each other at $P$ as desired.$\blacksquare$

Let $K$ be where the external bisector of $\angle A$ meets line $BC$,and let $P$ be the foot from $I$ to line $KI_A$. We claim the tangency point is $P$. Claim: $P$ lies on $(AID)$. Proof: The antipode of $I$ in $(AID)$ is $K$ – since $\angle IPK = 90^{\circ}$, $P$ also belongs to $(AIDK)$. Claim: $P$ lies on $(EI_AF)$. Proof: It would suffice to show that $PBDE$ is cyclic, since then it follows that $P$ is the Miquel point of concave quadrilateral $BCFE$. This cyclic quadrilateral follows from \[\angle PBE = \angle PII_A = 180^{\circ} - \angle PIA = 180^{\circ} - \angle PDA = \angle PDE.\] Claim: $(AID)$ is tangent to $(EI_AF)$. Proof: We have

Eyed wrote: Let $ABC$ be a triangle with $AB < AC$, incenter $I$, and $A$ excenter $I_{A}$. The incircle meets $BC$ at $D$. Define $E = AD\cap BI_{A}$, $F = AD\cap CI_{A}$. Show that the circumcircle of $\triangle AID$ and $\triangle I_{A}EF$ are tangent to each other Let $\Gamma$ be the circumcircle of $\triangle ABC$ and $N$ be the midpoint of $\overarc{BAC}$ of $\Gamma$ $\hspace{0.4cm}$$W$ be the intersection of line $NI_a$ and $\Gamma$ ($W \neq N$) $\textbf{First}$, we'll prove that $\angle WAC = \angle DAB$ : $\, \,$Let $M$, $L$ be the intersections of line $I_aA$, $I_aC$ with $\Gamma$ respectively ($M \neq A$, $L \neq C$) $\, \,$Cuz $\triangle I_aWA \sim \triangle IaMN$, $\triangle I_aWC \sim \triangle I_aLN$, $\triangle I_aAC \sim \triangle I_aLM$ we have that : $$ \dfrac{CW}{AW} = \dfrac{NL. \dfrac{I_aW}{I_aL}}{MN . \dfrac{I_aW}{I_aM}} = \dfrac{NL}{MN}.\dfrac{I_aM}{I_aL} = \sin \angle NML .\dfrac{I_aC}{I_aA}$$$\, \,$In the order hand, we ez to have that : $\angle NML = \angle MAC + \angle ICA = 90^\circ - \angle IBD$ and $\triangle AIB \sim \triangle ACI_a$ $\, \,$Therefore, $$\dfrac{CW}{AW} = \cos \angle IBD. \dfrac{AI}{BI} = \dfrac{BD}{AB}$$$\, \,$Combine with $\angle ABC = \angle AWC$, we see that : $\triangle ABD \sim \triangle AWC$. Lead to $\angle BAD = \angle WAC$ $(*)$ $\textbf{Next}$, let $V$ be the intersection of lines $AN$, $BC$ and $T$ be the interserction of lines $IN$, $VI_a$ $\hspace{1.4cm}$$\omega$ be the circle go through $T$ and tangent with $NI_a$ at $NI_a$, we'll prove that $\omega$ tangent with circle $(AID)$ : $\, \,$Since $\angle VAI = \angle VDI = 90^\circ$, we have that $A$, $D$ are lie on the circle with diameter $IV$ $\, \,$Combine with $\angle BAD = \angle WAC$ at $(*)$, we have that $\angle IVD = \angle IAD = \angle MAW = \angle MNW$ $\, \,$But $MN \perp VD$, so $NW \perp IV$. Lead to $I$ is the orthocenter of $\triangle NI_aV$, $NI \perp VI_a$ $\, \,$So $T$ lies on circle $(AID)$. Combine with $NI_a$ tangent $\omega$ at $I_a$ and $IV \perp NI_a$, it's clearly that $(AID)$ tangle $\omega$ at $T$ $\textbf{In the final}$, we just need to prove that $E$, $F$ all line on $\omega$ : $\, \,$Let $Y$ be the intersection of line $BI$ and $(AID)$ ($Y \neq I$); $E'$ be the intersection of lines $AD$, $YT$ $\, \,$Use Pascal theorem for $\binom{A\ V\ Y}{T\ I\ D}$, we have that $E'$, $B$, $I_a$ are colinear, which means $E$, $T$, $Y$ are colinear $\, \,$Therefore, $\angle ETI_a = \angle VTY = \angle VIB = \angle BI_aN$ (Cuz $IB \perp BI_a$, $IV \perp NI_a$ $\, \,$Combine with $NI_a$ tangent with $\omega$ at $I_a$, we see that $E$ lies on $\omega$. Similar, $F$ lie on $\omega$, done

Easy for a g6

How is this a G6

oops I solved this problem again -- but here's a solution by pure angle chasing (without adding any points (!!)) Claim: The Miquel points of convex quadrilateral $CI_AED$ and self-intersecting $EFI_AB$ coincide at a point $P$ on $(II_A)$. Proof: Let $P$ be the Miquel point of $EFI_AB$. Then $\measuredangle PI_AC = \measuredangle PED = \measuredangle PBC$, so $P$ lies on $(CI_ABI)$. Then $\measuredangle DFP = \measuredangle BI_AP = \measuredangle DCP$ so $P$ lies on $(CDF)$, which implies the result. $\blacksquare$ Denote $\theta = \angle BAD$. Claim: $P$ lies on $(AID)$. Proof: More angle chasing. \[\angle DPI = \angle CPI - \angle CPD = \frac B2 - \angle CFD = \frac A2 - \theta = \angle IAD. \ \blacksquare\] So now we show that $(AID)$ and $(I_AEF)$ are tangent at $P$. This ends up being yet another angle chase: \begin{align*} \angle DIP + \angle EI_AP &= 180^\circ - \angle DPI-\angle IDP \\ &= 90^\circ - \angle DAI - \angle BDP + \angle DCP \\ &= 90^\circ - \left(\frac A2 - \theta\right) - \angle CFD \\ &= 90^\circ - \left(\frac A2 - \theta\right) - \left(\theta+B+\frac C2 - 90^\circ\right) \\ &= 90^\circ - \frac B2 = \angle EBD = \angle EPD. \end{align*}Thus done.

The fastest way to calculate should be using Casey Theorem on $\odot (I_AEF),$ we only need $$I_AF\cdot\sqrt{I_AA\cdot I_AI}+I_AE\cdot\sqrt{FA\cdot FD}=EF\sqrt{I_AA\cdot I_AI}.$$Clearly all segments are easy to calculate.$\Box$