We can angle chase as well as directly use moving points. Soln 1. Let $DE\cap AC=X$, $DF\cap AB=Y$. Now, observe that $BD=BY, CD=CX$, thus we have that $IX=ID=IY$ and $I$ is on angle bisector of $\angle XAY$. Thus, $XIAY$ is cyclic. Also, $\measuredangle IFY=\measuredangle IAF$ so $IAFY$ is cyclic. Similarly, we get $IAEX$ cyclic. Thus $I, A, E, F, X$ and $Y$ are concyclic. Now, let $(DEF)\cap BC=A'$. Observe that $\angle EAF=EDF$, so $(DEF)$ is reflection of $(IAEFXY)$ in $EF$. Thus, but now $\measuredangle FEA'=\measuredangle FDB=\measuredangle BYF=\measuredangle AYF=-\measuredangle FYA=-\measuredangle FEA$. Similarly, $\measuredangle EFA'=-\measuredangle EFA$. Thus, $A'$ is reflection of $A$ in $EF$ and we are done. MMP Sketch. Move $D$ linearly along $BC$. $X,Y$ have degree $1$ so reflection of $A$ in $XY$ is degree $2$. So, we only need $3$ cases but its easy to prove when $D=B,C$ or $DI\perp BI$ or $DI\perp CI$ and we get $4$ cases so we are done. $D=AI\cap BC$ and $DI\perp BC$ are also fairly direct cases.

Simson Line FTW Let $D'$ be the reflection of $D$ in $BI$. Then $D' \in AB$. By easy angle chasing, we can get $$\angle D'EI = \frac12 \angle BAC = \angle D'AI$$$\implies$ $D',E,A,I$ are concyclic $\implies$ $\angle EAI = \angle ID'D = \angle IDE$. But, $\angle IDE = \angle IFE$ because $I$ is the orthocenter of $\triangle DEF$ $\implies$ $\angle EAI = \angle EFI$ $\implies$ $A,E,F,I$ are concyclic. Therefore by Simson's line theorem, reflections of $A$ in $IE,IF,EF$ are collinear. But reflections of $A$ in $IE$ and $IF$ lie on $BC$, and so we are done. $\blacksquare$

anser wrote: Let $ABC$ be a triangle with incenter $I$ and let $D$ be an arbitrary point on the side $BC$. Let the line through $D$ perpendicular to $BI$ intersect $CI$ at $E$. Let the line through $D$ perpendicular to $CI$ intersect $BI$ at $F$. Prove that the reflection of $A$ across the line $EF$ lines on the line $BC$. We set up cartesian coordinate system with $BC$ as x-axis and y-axis the perpendicular from $I$ on $BC$ and unit lenght so that $I \leftrightarrow (0,1)$. Let $B \leftrightarrow (b,1) , C \leftrightarrow (c,0) , D \leftrightarrow (d,0)$ Slope of $CI$ is $\frac{-1}{c}$ and hence of $DF$ is $c$. So equation of $\overleftrightarrow{DF}$ is $y = cx - cd$ And equation of $\overleftrightarrow{BI}$ is $by+x-b=0$ Solving we get $$ F \leftrightarrow \left(\dfrac{bcd+b}{bc+1} , \dfrac {cb-cd}{bc+1}\right)$$Similarly $$ E \leftrightarrow \left(\dfrac{bcd+c}{bc+1} , \dfrac {cb-bd}{bc+1}\right)$$ Hence equation of $\overleftrightarrow{EF}$ is $ L(x,y) = (y-xd)(bc+1) +bcd^2+cd+bd-bc = 0$ Let $A\leftrightarrow (t,s)$, then $y-$coordinate of reflection of $A$ in $EF$ is $s - \dfrac{2L(s,t)}{(bc+1)(d^2+1)}$ So we need this to be $0$ , that is $s(d^2+1)(bc+1) - 2L(s,t) = 0$

Now we find $A$ , $\overleftrightarrow{AB}$ has equation $2bx-2b^2+(b^2-1)y = 0$ And $\overleftrightarrow{AC}$ has equation $2cx-2c^2+(c^2-1)y = 0$ So $$A \leftrightarrow \left(\frac{b+c}{bc+1} ,\frac{2bc}{bc+1}\right) $$ Hence we need to show $$2bc(d^2+1) - 2 [(2 bc - (b+c)d)+bcd^2+d(b+c) -bc ] = 0$$$$\iff 2bcd^2 + 2bc - 2 [ bc + bcd^2] = 0$$ Which is true QED.

Also not very painful with complex bash with $I$ at origin (ofc some symmetry assumed)

[asy][asy] import olympiad; unitsize(1.5inch); pair A, B, C, D, Y, Z, P, Q, E, F, R, I, p, q, r; A = dir(102); B = dir(195); C = dir(345); D = 0.8B + 0.2C; I = incenter(A, B, C); Z = 2*foot(D, B, I) - D; Y = 2*foot(D, C, I) - D; E = extension(D, Z, C, I); F = extension(B, I, D, Y); P = foot(A, B, I); Q = foot(A, C, I); R = foot(A, E, F); r = 2R - A; q = 2Q - A; p = 2P-A; draw(A--B--C--cycle); draw(circumcircle(A, B, C), dashed+lightblue); draw(circumcircle(A, E, F), dotted); draw(C--E, purple); draw(B--P, purple); draw(E--D--Y, magenta); draw(E--F, lightred); draw(B--r); draw(P--R, dashed); draw(A--r, dotted+mediumred); draw(A--q, dotted+mediumred); draw(A--p, dotted+mediumred); dot("$A$", A, dir(90)); dot("$B$", B, dir(225)); dot("$C$", C, dir(C)); dot("$D$", D, dir(250)); dot("$Z$", Z, dir(200)); dot("$Y$", Y, dir(70)); dot("$F$", F, dir(300)); dot("$E$", E, dir(E)); dot("$I$", I, dir(120)); dot("$P$", P, dir(60)); dot("$Q$", Q, dir(290)); dot("$R$", R, dir(180)); dot(p); dot(q); dot(r); [/asy][/asy] Let $Y, Z$ be the reflections of $D$ over $CI, BI$, which lie on $AC, AB$, respectively. Since $DF\perp IE$ and $DE\perp IF$, $D$ is the orthocenter of $IEF$ and $Y, Z$ lie on $(IEF)$. $A$ also lies on $(IFZEY)$ because $\angle AIE = 90 - \tfrac{\angle B}{2} = \angle BZD = \angle AZE$. So if $P, Q, R$ are the feet from $A$ to $FI, EI, EF$, then $P, Q, R$ are collinear (Simson line). $P$ and $Q$ lie on the $A$-midline of $\triangle ABC$ because the reflections of $A$ over the $B$ and $C$-angle bisectors lie on $BC$. So the reflection of $A$ over $R$ lies on $BC$ as well.

EDIT: I should mention that, despite this solution's short length, this took me almost an hour to find. Granted, I was drawing freehand, but still. Tricky problem. Note that $\angle IBD = \angle IBA$ and \[ \angle BFD = \angle BIC - 90^\circ = \angle BAI, \]so $\triangle BDF\sim\triangle BIA$, whence $\angle BFA = \angle BDI$ by spiral similarity. Analogously, $\angle CEA = \angle CDI$, so quadrilateral $AEIF$ is cyclic. This means the perpendiculars from $A$ to the sides of $\triangle IEF$ are collinear. But this line is precisely the $A$-midline of $\triangle ABC$. Yay.

Let $X=DE\cap AB$ and $Y=DF\cap AC$ It is easy to see that $I$ is the circumcenter of $(XYD)$. In particular, $IX=IY\implies (AXIY)$-cyclic. Also, easy angle chase would show that $(AIEX), (AIFY)$ are cyclic. So, $(AEXIFY)$ is a cyclic pentagon. Now, let $A_1$ be the second intersection of $(DEF)$ with line $BC$. Then, $$\measuredangle FEA=\measuredangle FIA=\measuredangle BIA=\measuredangle BDF=\measuredangle A_1DF=A_1EF$$Similarly, $\measuredangle EFA=\measuredangle A_1FE$. Hence, $A_1$ is indeed the reflection of $A$ across $EF$.

Rg230403 wrote: We can angle chase as well as directly use moving points. Soln 1. Let $DE\cap AC=X$, $DF\cap AB=Y$. Now, observe that $BD=BY, CD=CX$, thus we have that $IX=ID=IY$ and $I$ is on angle bisector of $\angle XAY$. Thus, $XIAY$ is cyclic. Also, $\measuredangle IFY=\measuredangle IAF$ so $IAFY$ is cyclic. Similarly, we get $IAEX$ cyclic. Thus $I, A, E, F, X$ and $Y$ are concyclic. Now, let $(DEF)\cap BC=A'$. Observe that $\angle EAF=EDF$, so $(DEF)$ is reflection of $(IAEFXY)$ in $EF$. Thus, but now $\measuredangle FEA'=\measuredangle FDB=\measuredangle BYF=\measuredangle AYF=-\measuredangle FYA=-\measuredangle FEA$. Similarly, $\measuredangle EFA'=-\measuredangle EFA$. Thus, $A'$ is reflection of $A$ in $EF$ and we are done. MMP Sketch. Move $D$ linearly along $BC$. $X,Y$ have degree $1$ so reflection of $A$ in $XY$ is degree $2$. So, we only need $3$ cases but its easy to prove when $D=B,C$ or $DI\perp BI$ or $DI\perp CI$ and we get $4$ cases so we are done. $D=AI\cap BC$ and $DI\perp BC$ are also fairly direct cases. I don't think the reflexion is degree 2, the naive way to approach it will give degree 4. ( Let $P$ be the projection of $A$ on $EF$, which has degree 4, then $A'$ is the image of $P$ under the homothety of center $A$ with ratio 2. ) To PM#12 @Rg230403 I agree that intrinsically it would be degree 2, but the main purpose of moving point is to derive the degree without going through the computation (where in this problem direct bashing is not too hard). All I want to say is that getting to degree 2 conclusion is non-trivial, at least I don't know how to do it. Anyway, you have enough cases for degree 4 which can be obtained in a more naive way. Cheers

Flash_Sloth wrote: Rg230403 wrote: We can angle chase as well as directly use moving points. Soln 1. Let $DE\cap AC=X$, $DF\cap AB=Y$. Now, observe that $BD=BY, CD=CX$, thus we have that $IX=ID=IY$ and $I$ is on angle bisector of $\angle XAY$. Thus, $XIAY$ is cyclic. Also, $\measuredangle IFY=\measuredangle IAF$ so $IAFY$ is cyclic. Similarly, we get $IAEX$ cyclic. Thus $I, A, E, F, X$ and $Y$ are concyclic. Now, let $(DEF)\cap BC=A'$. Observe that $\angle EAF=EDF$, so $(DEF)$ is reflection of $(IAEFXY)$ in $EF$. Thus, but now $\measuredangle FEA'=\measuredangle FDB=\measuredangle BYF=\measuredangle AYF=-\measuredangle FYA=-\measuredangle FEA$. Similarly, $\measuredangle EFA'=-\measuredangle EFA$. Thus, $A'$ is reflection of $A$ in $EF$ and we are done. MMP Sketch. Move $D$ linearly along $BC$. $X,Y$ have degree $1$ so reflection of $A$ in $XY$ is degree $2$. So, we only need $3$ cases but its easy to prove when $D=B,C$ or $DI\perp BI$ or $DI\perp CI$ and we get $4$ cases so we are done. $D=AI\cap BC$ and $DI\perp BC$ are also fairly direct cases. I don't think the reflexion is degree 2, the naive way to approach it will give degree 4. ( Let $P$ be the projection of $A$ on $EF$, which has degree 4, then $A'$ is the image of $P$ under the homothety of center $A$ with ratio 2. ) The naive approach doesn't work in this case you're write but something like BOBTHEGR8's bash actually shows it's degree 2. I'd seen this bash so I knew it's degree 2 but it felt too tiresome to write out so I just left it as a sketch.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -20.563185329610068, xmax = 12.377859343880997, ymin = -11.613203123031788, ymax = 8.91955566448047; /* image dimensions */ /* draw figures */ draw((-4.211484462424216,4.268485741937438)--(-9.744668905056475,-3.510842880377173), linewidth(0.8) + blue); draw((-9.744668905056475,-3.510842880377173)--(-1.1983642213868468,-5.976123077589549), linewidth(0.8) + blue); draw((-1.1983642213868468,-5.976123077589549)--(-4.211484462424216,4.268485741937438), linewidth(0.8) + blue); draw((-9.744668905056475,-3.510842880377173)--(-8.569075859890033,1.3562595870223788), linewidth(0.8) + blue); draw((-9.744668905056475,-3.510842880377173)--(-3.4423550674533905,-1.3108294790664943), linewidth(0.8) + blue); draw((-1.1983642213868468,-5.976123077589549)--(-8.569075859890033,1.3562595870223788), linewidth(0.8) + blue); draw((-1.1983642213868468,-5.976123077589549)--(-3.4423550674533905,-1.3108294790664943), linewidth(0.8) + blue); draw((-6.548193390901448,-4.432903124844963)--(-8.569075859890033,1.3562595870223788), linewidth(0.8) + blue); draw((-6.548193390901448,-4.432903124844963)--(-3.4423550674533905,-1.3108294790664943), linewidth(0.8) + blue); draw((-8.569075859890033,1.3562595870223788)--(-3.4423550674533905,-1.3108294790664943), linewidth(0.8) + blue); draw((-4.211484462424216,4.268485741937438)--(-8.569075859890033,1.3562595870223788), linewidth(0.8) + blue); draw((-4.211484462424216,4.268485741937438)--(-3.4423550674533905,-1.3108294790664943), linewidth(0.8) + blue); draw((-8.45246494259377,-3.8835940233952604)--(-3.4423550674533905,-1.3108294790664943), linewidth(0.8) + blue); draw((-8.45246494259377,-3.8835940233952604)--(-8.569075859890033,1.3562595870223788), linewidth(0.8) + blue); draw((-3.4423550674533905,-1.3108294790664943)--(-2.769455569060166,-0.6344124955002953), linewidth(0.8) + blue); /* dots and labels */ dot((-4.211484462424216,4.268485741937438),dotstyle); label("$A$", (-4.055733570584836,4.635742680035466), NE * labelscalefactor); dot((-9.744668905056475,-3.510842880377173),dotstyle); label("$B$", (-9.595146912539612,-3.1563654209808756), NE * labelscalefactor); dot((-1.1983642213868468,-5.976123077589549),dotstyle); label("$C$", (-1.0644503659292572,-5.593707291440964), NE * labelscalefactor); dot((-5.252763527080084,-1.942807348727734),linewidth(4pt) + dotstyle); label("$I$", (-5.089757394416394,-1.6422591075132453), NE * labelscalefactor); dot((-6.548193390901448,-4.432903124844963),dotstyle); label("$D$", (-6.382287174205842,-4.0796009779733335), NE * labelscalefactor); dot((-8.569075859890033,1.3562595870223788),linewidth(4pt) + dotstyle); label("$E$", (-8.41340539958926,1.6444594753799036), NE * labelscalefactor); dot((-3.4423550674533905,-1.3108294790664943),linewidth(4pt) + dotstyle); label("$F$", (-3.686439347787851,-2.085412174869625), NE * labelscalefactor); dot((-8.45246494259377,-3.8835940233952604),dotstyle); label("$A'$", (-8.302617132750164,-3.5256596437778587), NE * labelscalefactor); dot((-7.816422833541189,-0.799843453098276),linewidth(4pt) + dotstyle); label("$D_1$", (-7.67481695399529,-0.49744701684259796), NE * labelscalefactor); dot((-2.769455569060166,-0.6344124955002953),linewidth(4pt) + dotstyle); label("$D_{2}$", (-2.6154861016765945,-0.3497293277238048), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $D_1$ be the intersection of $DE$ with $AB$, obviously we have that $D_1$ is the relfection of $D$ across $BI$ this implies that $D_1$ is on $AB$. Now let $D_2$ be the intersect3ion of $DF$ with $AC$, similarly we have that $D_2$ is on $AC$. Now by a quick angle calculation we have that $\angle D_1EF = \angle D_1AI = \frac{1}{2} \angle A$, thus we have that $D_1EAI$ is cyclic, similarily we have that $AIFD_2$ is also cyclic. Since we have that $ID_1=ID=ID_2$ and since we have that: $$\measuredangle ID_1A \equiv \measuredangle BD_1I \equiv \measuredangle IDB \equiv \measuredangle CDI \equiv \measuredangle ID_2C \equiv \measuredangle ID_2A$$we have that $ID_1AD_2$ is also cyclic. Meaning that $AED_1IFD_2$ is a cyclic hexagon. Now just use the Simson line of $A$ wrt $\triangle IEF$ with a homothethy factor of $2$ to get that $A' \in \overline{BC}$, where $A'$ is the reflection of $A$ across $EF$.

Let $E_1\in DE\cap DI$,$E_2\in DE\cap BA$,$F_1\in DF\cap CI$ and $F_2\in DF\cap CA$. We have that $\Delta DBE_2$ is isosceles($\overline{BE_1}$ is both an altitude and an angle bisector), which implies that $BI$ is the perpendicular bisector of $\overline{DE_2}$. Similarly, we obtain that $CI$ is the perpendicular bisector of $\overline{DF_2}$. Hence, $I$ is the circumcenter of $\Delta DE_2F_2$. Now, $\angle F_2IE_2=2\angle F_2DE=2(-\angle E_2DB-\angle CDF_2)=2(-(90-\angle DBE_1)-(90-\angle F_1CD))=2(\angle DBI+\angle ICD)=2(90-\angle BAI)=-2\angle BAI=-\angle BAC=-\angle E_2AF_2$. This means that $IE_2AF_2$ is cyclic. Also, we have $\angle E_2EI=\angle DEF_1=90-\angle F_1DE=\angle BAI=\angle E_2AI$ and $\angle IFD=\angle E_1FD=90-\angle FDE_1=\angle BAI=\angle IAF_2$. This means that $IE_2EAF_2F$ is cyclic. Let $Y$ be the foot of the perpendicular from $A$ to $EF$ and let $X\in AY\cap BC$. We have $\angle XAF=\angle YAF=90-\angle AFY=90-\angle AFE=90-\angle AE_2E=90-\angle BE_2E_1=\angle E_1BE_2=\angle DBE_1=\angle XBF$. This means that $XBAF$ is cyclic. Now, $\angle FXA=\angle FBA=\angle XBF=\angle XAF \Rightarrow \Delta AFX$ is isosceles. Since $\overline{FY}$ is an altitude of this triangle, it is also a median, which means that $AY=YX$, concluding that $X$ is the reflection of $A$ over $EF$.

Lemma: $AEIF$ is cyclic. Proof: Let the reflection of $A$ over $FI$ be $A'$. Notice that since $\angle ABF = \angle CBF$, $A'$ lies on $BC$. It also follows that $\angle BA'I = \angle BAI = \angle BFD$. Thus, $IDA'F$ is cyclic. We also see that $I$ is the orthocenter of $\triangle DEF$. Thus, we see that $\angle FEI = \angle FDI = \angle FA'I = \angle FAI$. $\square$ Then, let $X$, $Y$, $Z$ be the perpendiculars from $A$ to $EI$, $FI$, $EF$ respectively, and let $M$ and $N$ be the midpoints of $AB$ and $AC$. Iran Lemma tells us that $X$ and $Y$ lie on $MN$, and by Simson Line we see that $Z$ also lies on $MN$. The result then follows. $\blacksquare$

zuss77 wrote:

I got the same solution to but i did it respect to $A$. I dont wnat to do a double post of the same solution

Well probably this is the same as most of the solutions here. Let the measures angles of the triangle be $A$, $B$ and $C$ (idk how to write alpha, beta, gamma) We have that $<FDC=90-C/2$ and $<FBC=B/2$, so $<BFD=A/2$. This motivates to let $DF$ to intersect $AI$ at $X$. Now we see that $<BFX=<BAX$, so $BXFA$ is cyclic. Thus $<AXF=<IBA=<IBD$, so $BIDX$ is cyclic as well. Therefore $<BDI=<BXI=<AFI$. So we arrived at $<BDI=<AFI$. Similarly we get that $<AEI=<IDC$, and the last two equalities prove that $AEIF$ is cyclic. Now let $A'$ be the intersection point of $BC$ and $(EDF)$. We prove that $A'$ is the reflection of $A$ across $EF$. Note that $<A'EF=<A'DF=90-C/2=<AIF=<AEF$. Similarly $<AFE=<A'FE$, which proves the symmetry. $\blacksquare$

The map $E \mapsto D \mapsto F$ is projective and $E,F$ are moving on fixed lines hence line $EF$ is tangent to a fixed conic $\Gamma$. By considering $D=\infty_{BC}$ we get $EF$ is the line at infinity which is tangent to $\Gamma$ hence $\Gamma$ is a parabola. Let $I_{A},I_{B},I_{C}$ be the excentres opposite $A,B,C$. From $D=B,C$ we get $I_{A}I_{B}$, $I_{A}I_{C}$ tangent to $\Gamma$ and from choosing $D$ so $I=E,F$ we get $BI_{B}$, $CI_{C}$ tangent to $\Gamma$. These 4 tangents uniquely determine $\Gamma$ and from here we see this means $\Gamma$ has focus $A$ and directrix $BC$. Now it's well-known that the reflection of the focus in the tangent of a parabola lies on the directrix which means the reflection of $A$ in $EF$ will lie on $BC$.

We begin as follows: Claim: $AEIF$ is cyclic. Proof. Let $X = \overline{CA} \cap \overline{DF}$. Then \begin{align*} \measuredangle FXA &= \measuredangle DXC = \measuredangle CDX = 90^{\circ} - \frac{1}{2} C = \measuredangle AIB = \measuredangle FIA \\ \measuredangle EXF &= \measuredangle EXD = \measuredangle XDE = \measuredangle FDE = \measuredangle EIF \end{align*}which implies $AFXI$ and $FXIE$ are cyclic, respectively. $\blacksquare$ [asy][asy] pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = 0.7*C+0.3*B; pair I = incenter(A, B, C); pair E = extension(C, I, D, foot(D, B, I)); pair F = extension(B, I, D, foot(D, C, I)); filldraw(A--B--C--cycle, invisible, deepcyan); draw(B--F, lightblue); draw(C--I--A, lightblue); filldraw(D--E--F--cycle, invisible, red); filldraw(circumcircle(A, E, F), invisible, deepgreen+dashed); pair Y = extension(D, E, A, B); pair X = extension(D, F, A, C); draw(D--Y, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(270)); dot("$I$", I, dir(250)); dot("$E$", E, dir(225)); dot("$F$", F, dir(F)); dot(Y); dot("$X$", X, dir(X)); /* TSQ Source: A = dir 110 B = dir 210 C = dir 330 D = 0.7*C+0.3*B R270 I = incenter A B C R250 E = extension C I D foot D B I R225 F = extension B I D foot D C I A--B--C--cycle 0.1 lightcyan / deepcyan B--F lightblue C--I--A lightblue D--E--F--cycle 0.1 lightred / red circumcircle A E F 0.1 yellow / deepgreen dashed Y .= extension D E A B X = extension D F A C D--Y red */ [/asy][/asy] Now, the dilation of the Simson line of $A$ to $\triangle IEF$ should be collinear, but the reflections of $A$ about lines $BI$ and $CI$ lie on line $BC$ by definition. This solves the problem.

Let $DE$ and $DF$ intersect $AB$ and $AC$ again at $K$ and $L$, respectively. Claim: $AKIL$ is cyclic. Proof: Note that $I$ is the circumcenter of $\triangle DKL$, so $IK = IL$. But $I$ lies on a bisector (either internal or external depending on the configuration) of $\angle KAL$, so $I$ is the midpoint of one of the two arcs $\overset{\Huge\frown}{KL}$ of $(AKL)$, as desired. Claim: $AEIF$ is cyclic. Proof: We have $$\measuredangle IFL = \measuredangle IFY = 90^\circ + \measuredangle FIY = 90^\circ + \measuredangle FIC = \frac{1}{2}\measuredangle BAC = \measuredangle IAL$$ so $F$ lies on $(AIL)$. Similarly, $E$ lies on $(AIK)$, so we are done by the previous claim. Now, let $X$ and $Y$ be the feet from $A$ to lines $BI$ and $CI$, respectively, and $M$ and $N$ the midpoints of $AB$ and $AC$, respectively. We have $$\measuredangle XMA = 2\measuredangle XBA = \measuredangle CBA$$ so $X$ lies on line $MN$. Similarly, $Y$ lies on line $MN$. But the foot from $A$ to $EF$ lies on line $XY$ by simson line, so we are done.

Let $P=DE\cap AB$. Notice that $I$ is the orthocenter of $\triangle DEF$. So \[ 90^\circ+\angle A/2=\angle BIC=180^\circ-\angle EDF\implies \angle EDF=90^\circ-\angle A/2 \]Now \[ \angle PEI=90^\circ-\angle BIE=90^\circ-\angle EDF=\angle A/2 \]Means $AEPI$ is cyclic. But $\angle PFI=\angle IFD=\angle A/2$, which gives $F\in (AEPI)$. Let $K=BC\cap (DEF)$. $K$ is the reflection of $A$ wrt $EF$ because $\angle EKF=\angle EAF=90^\circ-\angle A/2$ and $\angle KEF=\angle CDF=\angle AEF$. $\blacksquare$ Alternatively since $AEIF$ is cyclic and the reflection of $A$ wrt $IE$ and $IF$ lies on $BC$ therefore $BC$ is the Steiner line of $A$ wrt $(DEF)$, meaning reflection of $A$ wrt $EF$ lies on the Steiner line $BC$. $\blacksquare$

Simple angle chasing

Nice geometry. Another angle chasing solution. Solution: Let $X = DF \cap AC$ and $A' = \odot(DEF) \cap BC \ne D$. We will show that $A'$ is indeed the reflection of $A$ across $A$ across $EF$ which would be enough.

Claim: Points $E, I, X, F$ and $A$ are concyclic. Proof: First of all observe that $I$ is the orthocenter of $\triangle DEF$. Also, $\overline{ID} = \overline{IX}$ because $X$ is indeed the reflection of $D$ across the $C$ angle bisector of $\triangle ABC$. We now show $X \in \odot(EIF)$. \[ \measuredangle IXD = \measuredangle XDI = \measuredangle IEF \]At last we will now show $A \in \odot(EIGX)$. \[ \measuredangle IFX = 90^\circ - \measuredangle CIF = 90^\circ + \measuredangle BIC = \measuredangle IAX\]hence we are done with the claim. $\square$ To finish it off, it suffices to show $\measuredangle A'EF = \measuredangle FEA$ which would conclude $A'$ is reflection of $A$ across the desired line as this would even give $\triangle AEF \cong \triangle A'EF$ by AAS congruency postulate. \[ \measuredangle A'EF = \measuredangle A'DF = \measuredangle DXC = \measuredangle FXA = \measuredangle FEA \]so we are finally finished with the solution. $\blacksquare$

Cool problem. Won't deal with config issues for the time being. Set $A'$ to be the reflection of $A$ across $EF.$ Let $\beta = \frac{\angle ABC}{2}.$ Also let $X = DE \cap AB$ and $Y = DF \cap BC.$ We will show that $A,X,E,I,F,Y$ are concyclic. First of all, note that $BD = BX$ by the perpendicularity condition. Hence $\angle AXE = 180^\circ - \angle BXD = 90^\circ + \beta.$ However, $\angle AIC = 90^\circ + \beta,$ so $\angle AIE = 90^\circ - \beta.$ This implies that $A,X,E,I$ are concyclic. Similarly, $A,Y,F,I$ are concyclic. On top of that, $\angle XAI = \angle IAH$ and $XI = IY = ID$ by reflections, so by a phantom points argument, $A,X,I,Y$ are concyclic. These 3 sets of concyclic points imply the claim. In particular, we have $A',D,F,E$ are concyclic, so $$\angle EDA' = \angle EFA' = \angle AFE = 180^\circ - \angle AXE = \angle BXE = \angle EDB,$$as desired.

Amazing problem By angle chasing we have $\angle DEC = 90 + \frac{\angle A}{2}$, hence due to AA similarity, $\triangle EDC \sim \triangle AIC$, so it follows that $\angle IDC = \angle AEC$. Similarly, $\angle IDB = \angle AFB$, hence \[ \angle AEC + \angle AFB = \angle IDC + \angle IDB = 180^{\circ}, \]so $AEIF$ is cyclic. Let $X$ and $Y$ be the feet from $A$ to $\overline{CE}$ and $\overline{BF}$. By Simson lines, the foot from $A$ to $\overline{EF}$ lies on $\overline{XY}$. In fact, $\overline{XY}$ is the midline of $\triangle ABC$ since extending $AX$ and $AY$ to hit $\overline{BC}$ gives us isosceles triangles. Hence, the foot from $A$ to $\overline{EF}$ lies on the midline of $\triangle ABC$, so the reflection lies on $\overline{BC}$.

Let $X=DF\cap AC$ and $Y=DE\cap AB$. Claim: $AYEIXF$ is cyclic. Proof. Note that $I$ is the orthocenter of $DEF$ and circumcenter of $DXY$. Thus as $IX=IY$ and $AI$ bisects $\angle YAX$ we have that $IAYX$ cyclic. However, also we have that \[\measuredangle EIF=\measuredangle FDE=\measuredangle XDE=\measuredangle EXD=\measuredangle EXF\]and similarly $\measuredangle EIF=\measuredangle EYF$ implying $YEIXF$ cyclic and $A$ lies on this circle. $\blacksquare$ Now define $A'$ to be the second interscetion of $BC$ and $(DEF)$. First, observe that $(DEF)$ and $(AEF)$ are reflections about line $EF$. Then, we have \[\measuredangle AFE=\measuredangle AYE=\measuredangle BYD=\measuredangle YDB=\measuredangle EDA'=\measuredangle EFA'\]implying the result.

what is mmp Use coordinates with $A=(0,1)$ and $\overline{BC}$ as the $x$-axis. Move $D:=(t,0)$ linearly; clearly $E$ and $F$ move linearly as well, so they can be parametrized as $E=(f_1(t),g_1(t)),F=(f_2(t),g_2(t))$. Furthermore, similar triangles/law of sines implies that the leading coefficients of $f_1$ and $f_2$ are the same: in particular, they're equal to $\tfrac{\cos \tfrac{B}{2} \cos \tfrac{C}{2}}{\sin \tfrac{A}{2}}$. The equation of line $\overline{EF}$ is $$y=\frac{f_1-f_2}{g_1-g_2}x-\frac{f_1g_2-f_2g_1}{g_1-g_2}.$$Now the projection of $(0,1)$ onto the line $y=px+q$ turns out to have $y$-coordinate $\tfrac{p^2+q}{p^2+1}$, which applied to $\overline{EF}$ is a rational function with numerator and denominator quadratic. Thus to show that the $y$-coordinate is always $\tfrac{1}{2}$ it suffices to check three cases. When $D=B,C$, $\overline{EF}$ becomes an external angle bisector, so the conclusion is obvious. When $D$ is the $A$-intouch point, the iran lemma implies that $\overline{EF}$ is the $A$-midline, so the conclusion is obvious as well. $\blacksquare$

ok we use synthetic brain Let $P$ and $Q$ be the reflections of $D$ over $\overline{BI}$ and $\overline{CI}$ (so $D,E,P$ are collinear, as are $D,F,Q$), which lie on $\overline{AB}$ and $\overline{AC}$ respectively. Then $\measuredangle AIE=90^\circ+\measuredangle ABI$ and $\measuredangle APE=\measuredangle BPE=90^\circ+\measuredangle ABI$, hence $AIEP$ is cyclic. Similarly, $AIFQ$ is cyclic. Finally, we have $\measuredangle PEQ=\measuredangle DEQ=2\measuredangle EDQ=\measuredangle BAC=\measuredangle PAQ$, since we can easily get $\angle BDE=90^\circ+\angle B/2$ and $\angle CDF=90^\circ+\angle C/2$, hence $\angle EDF=90^\circ+\angle A/2$. Thus $AEPQ$ is cyclic. This implies that $AIEFPQ$ is cyclic. Now let $A'$ be the second intersection of $(DEF)$ with $\overline{BC}$. We have $\measuredangle A'EF=\measuredangle CDQ=\measuredangle DQC=\measuredangle FQA=\measuredangle FEA$ and likewise $\measuredangle A'FE=\measuredangle EFA$, hence $A'$ is the reflection of $A$ over $\overline{BC}$ as desired.

Here is a sketch Observe there exist 2 spiral similarities AIC --> EDC BIA --> BDF Then just angle chase and we are done

Feels fairly tricky for problem 4, but maybe I'm just rusty at cyclic quadrilateral scouting. Let $P = \overline{ED} \cap \overline{AB}$ and $Q = \overline{DF} \cap \overline{AC}$. Claim: $EPFIQA$ is cyclic. Proof: Notice that $EPFI$ is cyclic because $\angle IEP = \frac 12 \angle A = \angle PFB$, and similarly $EQFI$ is also cyclic. Now note that $P$ and $Q$ are the reflections of $D$ over $\overline{BI}$ and $\overline{CI}$ respectively, so that $PI = ID = IQ$. It follows from the converse of Fact 5 that $A$ must lie on this circle too. $\blacksquare$ Now let $A'$ be the reflection of $A$ about $\overline{EF}$. By the claim, $\measuredangle EA'F = \measuredangle FAE = \measuredangle FPD = \measuredangle EDF$, so $EFDA'$ is also cyclic. But now $\measuredangle A'DF = \measuredangle A'EF = \measuredangle FEA = \measuredangle FPB = \measuredangle BDF$, so it follows that $A'$ lies on $\overline{BC}$, as needed.

Claim: Quadrilateral $AEIF$ is cyclic. Let $DE$ meet $AB$ at $E_1$ and $DF$ meet $AC$ at $F_1$. Note that $E_1$ and $D$ are reflections across $BI$. Thus, we have $$\measuredangle AE_1E = \measuredangle BE_1D = \measuredangle E_1BD + \measuredangle BDE_1 = \measuredangle AIC = \measuredangle AIE,$$so $AE_1EI$ is concyclic. Similarly, $AF_1FI$ is concyclic. However, notice that we also have $$\measuredangle AE_1I = \measuredangle BE_1I = \measuredangle IDB = \measuredangle IDC = \measuredangle CF_1I = \measuredangle AF_1I,$$so in fact $AE_1IF_1$ is also cyclic. Thus, $AE_1EIF_1F$ is conyclic. Now, let $D_0$ be the foot from $I$ onto $\overline{BC}$, $E_0$ be the foot from $A$ onto $\overline{CI}$, and $F_0$ be the foot from $A$ onto $\overline{BI}$. From Iran lemma, we find that $E_0F_0$ is the $A$-midline of $\triangle ABC$, so if $H_A$ is the foot from $A$ onto $\overline{BC}$, then $H_A$ is the reflection of $A$ onto $\overline{BC}$. Moreover, since $\measuredangle AE_0I = 90^{\circ} = \measuredangle AF_0I$, we know that $AE_0IF_0$ is also cyclic. Thus, $A$ must be the Miquel point of $EE_0F_0F$, so $A$ is the center of the spiral similarity taking $\overline{E_0F_0} \mapsto \overline{EF}$. If we let $A'$ be the image of $H_A$ under this spiral similarity, then $A'$ be the reflection of $A$ across $\overline{EF}$. However, we also note that $\overline{E_0E} \cap \overline{H_AA'}$ must lie on $(AE_0H_A)$ and $(AEA')$. But because $AE_0H_AC$ is concyclic from $\measuredangle AE_0C = 90^{\circ} = \measuredangle AH_AC$, we find that $C = \overline{E_0E} \cap \overline{H_AA'}$. Therefore $A'$ lies on $H_AC$, which finishes.