We choice $VN$ perpendicular bisector of $ AE $ and we see $AE$ perpendicular bisector of $VN$

math1181 wrote: we see $AE$ perpendicular bisector of $VN$ Well, that's the whole problem. Can you prove it?

Let $I$ denote the incenter, $J$ the $A$-excenter, and $L$ the midpoint of $\overline{AE}$. Denote by $\overline{IY}$, $\overline{IZ}$ the tangents from $I$ to the $A$-excircle. Note that lines $\overline{BC}$, $\overline{GF}$, $\overline{YZ}$ then concur at $H$ (unless $AB=AC$, but this case is obvious), as it's the radical center of cyclic hexagon $BICYJZ$, the circumcircle and the $A$-excircle. [asy][asy] size(12cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair M = dir(270); pair J = 2*M-I; pair D = foot(J, B, C); pair E = foot(A, B, C); pair F = IP(CP(J, D), unitcircle); pair G = OP(CP(J, D), unitcircle); draw(unitcircle, lightblue); draw(arc(J,abs(D-J),-30,210), lightblue); pair L = midpoint(A--E); pair V = extension(G, D, L, L+B-C); pair N = extension(F, D, L, L+B-C); draw(A--B--C--cycle, lightblue); draw(A--E, lightblue); draw(G--V, lightblue); draw(N--F, lightblue); draw(E--V--A--N--cycle, heavygreen); draw(V--N, heavygreen); pair H = extension(G, F, B, C); draw(C--H--G, lightblue); draw(circumcircle(B, I, C), lightred); pair Y = IP(circumcircle(B, I, C), circumcircle(D, F, G)); pair Z = OP(circumcircle(B, I, C), circumcircle(D, F, G)); pair T = -D+2*foot(J, D, I); draw(T--H, blue); draw(Z--H, blue); draw(L--T, blue); draw(Y--I--Z, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$M$", M, dir(M)); dot("$J$", J, dir(J)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(270)); dot("$G$", G, dir(270)); dot("$L$", L, dir(L)); dot("$V$", V, dir(V)); dot("$N$", N, dir(N)); dot("$H$", H, dir(H)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(Z)); dot("$T$", T, dir(T)); [/asy][/asy] Now let $\overline{HD}$ and $\overline{HT}$ be the tangents from $H$ to the $A$-excircle. It follows that $\overline{DT}$ is the symmedian of $\triangle DZY$, hence passes through $I = \overline{YY} \cap \overline{ZZ}$. Moreover, it's well known that $\overline{DI}$ passes through $L$, the midpoint of the $A$-altitude (for example by homothety). Finally, $(DT;FG) = -1$, hence project through $D$ onto the line through $L$ parallel to $\overline{BC}$ to finish.

v_Enhance wrote: hence project through $D$ onto the line through $L$ parallel to $\overline{BC}$ to finish. Can someone explain me this part? Why do we need that lines are parallel for that? And in general how do we project from circle to line?

It's the projection $D(DT;FG)$ from the circle, which then becomes the bundle $(\infty L; NV) = -1$ which is the desired result. (See chapter 9 of my textbook EGMO, or http://www.mit.edu/~alexrem/ProjectiveGeometry.pdf for details on projective geometry.)

v_Enhance wrote: It's the projection $D(DT;FG)$ from the circle, which then becomes the bundle $(\infty L; NV) = -1$ which is the desired result. (See chapter 9 of my textbook EGMO, or http://www.mit.edu/~alexrem/ProjectiveGeometry.pdf for details on projective geometry.) Oh, thanks, I totally forgot to consider projection of $D$ as $DD$...

A very different solution (with K6160): Consider the negative inversion at $D$ fixing $\odot(ABC)$; since line $BC$ is fixed, the excircle maps to a line parallel to $BC$. I claim that the perpendicular distance from this line to $D$ is $\tfrac{r}{2}$, where $r$ is the inradius. To prove this, let $r_a$ denote the $A$-exradius and $K$ denote the area of $\triangle ABC$; it suffices to show that $$\frac{r}{2}\cdot 2r_a=DB\cdot DC$$$$\Longrightarrow rr_a=(s-b)(s-c)$$$$\Longrightarrow K^2=s(s-a)(s-b)(s-c)$$which is true, as desired. Now, let $DF$ and $DG$ intersect $\odot(ABC)$ again at $Y$ and $X$, respectively. By our earlier observation, $XY$ is parallel to $BC$, and also has perpendicular distance of $\tfrac{r}{2}$ to $BC$. If $I$ is the incenter, it consequently follows that $XY$ bisects $ID$. But since the midpoint of the projection of $I$ onto $BC$ and $D$ is also the midpoint of $\overline{BC}$, it follows that the midpoint of $\overline{ID}$ also lies on the perpendicular bisector of $\overline{BC}$, and hence it is also the midpoint of $\overline{XY}$. Letting the perpendicular bisector of $\overline{AE}$ intersect $DF$ and $DG$ at $N$ and $V$, respectively, and dilating through $D$ gives the conclusion by the midpoint of altitude lemma.

See here. https://www.artofproblemsolving.com/community/c6h17323p118682 I just think that the main idea about this problem is the same as this one. We can finish this problem immediately if you know this.

My solution: It clearly suffices to show that $(DM,DE;DG,DF)=-1$ where M is the midpoint of $AE$. Projecting onto the excircle, if $MD\cap \text{excircle} = X$, it suffices to show that $DFGX$ is harmonic or what is equivalent, tangent at X to the excircle, BC, FG concur. By extraversion of ISL 2002 G7, $(BCX)$ is tangent to the excircle and thus by an application of POP, we are done.

Oh wow this is a very nice problem Using a few lemmas we can solve this problem. Let $M$ be the midpoint of $AE$, and the perpendicular bisector of $AE$ be $l$. It suffices to show that the distance between $l \cap DG$ and $l \cap DF$ is bisected by $E$. Let $ED$ hit the excircle again at $X$. Projecting at $D$ onto the excircle, since $l \parallel BC$, it suffices to show that $DGXF$ is harmonic. We want to use the midpoint of altitude lemma, or the fact that $I$ lies on $ED$. This, with the duality of orthocenters and excenters motivates us to draw $BICI_a$. Then radical axis and chasing poles and polars solves this problem as done above

Here's a solution using a powers bash: Let $r_A$ be the $A-$exradius and $h_A$ be the $A-$altitude. Take $N, V$ to be on the perpendicular bisector of $AE$ and $T$ the midpoint of $AE$. We want to show $NT=VT$. Note $DV = \frac{h_A}{2\sin \angle VDC}$ and $DG=2r_A\sin\angle BDG=2r_A\sin\angle VDC$. Thus $DV\cdot DG=h_A\cdot r_A$. Similarly $DN\cdot DF=h_A\cdot r_A$, so $(FGNV)$ is cyclic. Let $p(P, \omega)$ denote the power of $P$ with respect to a circle $\omega$. Then we define $f(P)=p(P, (ABC))-p(P, (FGNV))$ and $g(P)=p(P, (DFG))-p(P, (FGNV))$. Note that $f$ and $g$ are linear in $P$. Conveniently, $f(F)=f(G)=g(F)=g(G)=0$. Thus it follows that $f(N)=\frac{VG}{DG}f(D)$, $g(N)=\frac{VG}{DG}g(D)$ and similarly for $V$. Let $K$ the projection of the center of $(ABC)$ onto $NV$ and $L$ the projection of $D$ onto $NV$. Note that $p(N, (ABC))-p(V, (ABC))=NK^2-VK^2$ and $p(N, (DFG))-p(V, (DFG))=NL^2-VL^2$. But since $p(N, (FGNV))=0$ it follows $f(N)=p(N, (ABC))$ and $g(N)=p(N, (DFG))$ and similarly for $V$. Thus we have $$\frac{NK^2-VK^2}{NL^2-VL^2}=\frac{p(N, (ABC))-p(V, (ABC))}{p(N, (DFG))-p(V, (DFG))} = \frac{f(N)-f(V)}{g(N)-g(V)}=\frac{\left(\frac{VG}{DG}-\frac{NF}{DF}\right)f(D)}{\left(\frac{VG}{DG}-\frac{NF}{DF}\right)g(D)}=\frac{f(D)}{g(D)}$$ Write $d=NT-VT$. Note that $\frac{NK^2-VK^2}{NL^2-VL^2}=\frac{NK-VK}{NL-VL}=\frac{d+2KT}{d+2LT}$. Moreover, we can compute $f(D)=-(s-b)(s-c)+h_Ar_A$ and $g(D)=h_Ar_A$. By projecting $T, K, L$ onto $BC$, it can be shown that $\frac{KT}{LT}=\frac{-(s-b)(s-c)+h_Ar_A}{h_Ar_A}$, and thus $d=0$, as desired.

Lemma:Let $ID\cap \odot I_A=\{P\}$ than $\odot BPC$ tangents to $\odot I_a$ Proof: Let $\odot I_A$ touch AB,AC,BC in E,F,D.Let the E,F-midlines of triangles $\triangle BED$ and $\triangle CFD$ cut in $R$.Now $R$ is the radical center of $B,C,\odot I_A$ so $BR=RC$ but also $t(R,\odot I_A)\cdot BC=BR\cdot BD+CR\cdot CD$ where $t(X,\odot A)$ is the length of a tangent from $X$ to $\odot A$, so by Casey's $\odot BRC$ tangent to $\odot I_A$ but $R$ is the midpoint of $ID$ and so if $ID\cap \odot I_A=\{P\}$ than ID bisects $\angle BPC$ $\implies$ $P\in \odot BRC$ and so the conclusion follows. Let the perpendicular bisector of $AE$ cut $DG,DF$ in $V,N$ and let $L$ be the midpoint of $AE$ I'll prove that $L$ is the midpoint of $VN$.Notice that $PP\cap BC$ is on the radical axis of $\odot BAC,\odot I_A$ and so $PP,DD,GF$ are concurrent $\implies$ $DGPF$ is harmonic and hence $ID$ is the symmedian in $\triangle DGF$ so the rest is proving that $GF,VN$ are anti-parallel,: $$\angle DGF=\angle EDF=\angle NVD$$Hence the conclusion.

v_Enhance wrote: It's the projection $D(DT;FG)$ from the circle, which then becomes the bundle $(\infty L; NV) = -1$ which is the desired result. (See chapter 9 of my textbook EGMO, or http://www.mit.edu/~alexrem/ProjectiveGeometry.pdf for details on projective geometry.) Hi...The link does't work. would you correct it?

@above http://alexanderrem.weebly.com/math-competitions.html

v_Enhance wrote: .Finally, $(DT;FG) = -1$ Can you understand me pls. How???

Look at the tangents from $H$ to see that $DFTG$ is a harmonic quadrilateral.

Here's my updated solution from August 14th, 2019. [asy][asy] unitsize(1.2inches); pair A=dir(130); pair B=dir(215); pair C=dir(-35); pair I=incenter(A,B,C); pair E=foot(A,B,C); pair M=(A+E)/2; pair U=extension(A,I,B,C); pair L=circumcenter(B,I,C); pair IA=2*L-I; pair P=foot(IA,A,B); pair Q=foot(IA,A,C); pair D=foot(IA,B,C); pair T=extension(P,Q,B,C); pair R=(D+T)/2; pair F = intersectionpoints(circumcircle(A,B,C), circumcircle(P,D,Q))[0]; pair G = intersectionpoints(circumcircle(A,B,C), circumcircle(P,D,Q))[1]; pair X=2*foot(IA,D,I)-D; draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(B--P); draw(C--Q); draw(circumcircle(P,D,Q)); draw(A--E); draw(F--G); draw(G--R); draw(P--T); draw(C--T); draw(M--X); draw(A--IA); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,2*dir(90)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-90)); dot("$D$",D,dir(90)); dot("$E$",E,dir(45)); dot("$M$",M,dir(15)); dot("$P$",P,dir(180)); dot("$Q$",Q,dir(90)); dot("$T$",T,dir(90)); dot("$R$",R,dir(90)); dot("$I$",I,dir(I)); dot("$U$",U,dir(60)); dot("$L$",L,dir(-110)); dot("$I_A$",IA,dir(IA)); dot("$X$",X,dir(X)); [/asy][/asy] Letting $N=FD\cap\ell$ and $V=GD\cap\ell$ where $\ell$ is the $A$-midline, the problem asks us to show that $M$ is the midpoint of $NV$, or that $(NV;M\infty)=-1$. Projecting through $D$ onto the excircle, we simply want to show that $(DX;FG)=-1$ where $X=MD\cap\omega_A$, where $\omega_A$ is the $A$-excircle. Let $R=FG\cap BC$. The problem then reduces to showing that $RX$ is tangent to $\omega_A$, or that the polar of $R$ is $DX$. It's well known that $M,I,D$ collinear, so to show that the polar of $R$ is $DX$, all we have to show is that $I$ is on the polar of $R$, or that $R$ is on the polar of $I$. Let $P,Q$ be the tangency points of $\omega_A$ to the sides of the triangle, and let $T=PQ\cap BC$. Note that $(TD;BC)=-1$. However, we have that \[RD^2=RF\cdot RG=RB\cdot RC,\]so $R$ is the midpoint of $TD$. Let $R'$ be the intersection of the polar of $I$ with respect to $\omega_A$ with $BC$. Our goal now is to show that $R'$ is the midpoint of $TD$. Define the projective map from $AI$ to $BC$ by $X\mapsto\ell_X\cap BC$ where $\ell_X$ is the polar of $X$ with respect to $\omega_A$. Applying this map, we see that \[-1=(AU;II_A)=(TD;R'\infty),\]so $R'$ is the midpoint of $TD$, as desired.

Let $M$ be the midpoint of $\overline{AE}$, and $I$ be the incenter. Also, let the tangents to the excircle $\omega$ at $F,G$ meet at $K$. Lemma $D,K, I, M$ are collinear. Proof: By a well known lemma (see EGMO Chapter 4), we have that $D, I, M$ are colinear. Now we need to show that $D, K, I$ are colinear. Let $\Gamma = \omega_{ABC}, \tau = \omega_{BICI_A}$, where $I_A$ is the $A$-excenter. By Fact 5, $\tau$ exists and $\angle IBI_A = \angle ICI_A = 90$, so the radical axis of $\{ \omega, \tau \}$ is the polar of $I$ w.r.t $\omega$. Also, clearly $\overleftrightarrow{BC}$ is the polar of $D$ w.r.t $\omega$, and $\overleftrightarrow{FG}$ is the polar of $K$ w.r.t $\omega$. So the Polars of $I, D, K$ are the radical axises of $\{ \omega, \tau \}, \{\Gamma, \tau \}, \{ \omega, \Gamma \}$, so they are concurrent at the radical center of $\{ \omega, \tau, \Gamma \}$. By La-Hire, the conclusion follows. $\blacksquare$. Now, returning to the main problem, let $\ell$ be the line through $M$ parallel to $\overleftrightarrow{BC}$, and define $V = DG \cap \ell, N = DF \cap \ell$. As $$ -1 = (D, DK \cap \omega ; G, F) \overset{\text{Lemma}}{=} (D, DM \cap \omega; G, F) \overset{D}{=} (P_{\infty}^{\ell}, M; V, N)$$, so $VM = MN$. As $AM = ME$, the quadrilateral $EVAN$ is a rhombus and we're done.

yayups wrote: Sorry dumb question but, what size paper is that(And where can I get some)? Also, is large paper allowed on usamo?

Define $N$ on $\overline{DG}$ and $V$ on $\overline{DF}$ such that $NV$ bisects $AE$ at $M$ and a right angle. It remains to show that $MN = MV$. Let $X$ be the intersection of $\overline{MD}$ with the $A$-excircle. Projecting through $D$, it remains to show that quadrilateral $FDGX$ is harmonic. It is well-known that the incenter $I$ lies on $MD$. Claim: Let $U$ and $W$ be the intersection points of the $A$-excircle with $(BICI_A)$. Then $I$ is tangent to $A$-excircle at $U$ and $W$. Proof. Follows since $II_A$ is a diameter of $(BICI_A)$. $\blacksquare$ Thus, by radical axis on the $A$-excircle, $(BICI_A)$, and the circumcircle it follows that $\overline{BC}$, $\overline{FG}$, and $\overline{UW}$ concur at some point $R$. Since $R$ lies on the polar of $I$ wrt to the $A$-excircle, it follows that $I$ lies on the polar of $R$, which thus must be $\overline{DX}$. As such, harmonicity follows.

Let DG, DF hit the perpendicular bisector of AE at V,N. Define H=VN \cap AE as the midpoint of AE. Claim: If line DH hits the A-excircle at J, then (DJ;FG)=-1. Proof. Let the tangent from J to the A-excircle hit BC at K. It suffices to show K lies on line FG, the radical axis of (ABC) and the A-excircle. Thus we show that KD^2=KC\cdot KD which will imply that K is on the radical axis. Now, invert around \omega, the circle with center K passing through J. First, it is known that the incenter I of \triangle ABC lies on line H-D-J. Then, notice that I_A inverts to the midpoint M of DJ. But then, \angle DMI_A=90 ^{\circ} so \omega is orthogonal to the circle with diameter II_A. Fact 5 gives B,C lie on this circle proving the claim. \blacksquare Then, to finish -1=(FG;JD)\overset{D}{=}(NV;H\infty) showing H is the midpoint of VN, done.

stupid. i dislike it strongly Let $M$ be the midpoint of $\overline{AE}$ and let $\overline{DM}$ intersect the $A$-excircle again at $T$. Let $I$ and $I_A$ denote the incenter and $A$-excenter, respectively, of $\triangle ABC$. It suffices to prove that $DFXG$ is harmonic, whence projecting through $D$ yields the desired result. Let $N$ be the midpoint of $\overline{DX}$ and let $T$ be the intersection of the tangents to the $A$-excircle at $D$ and $X$. It suffices to show that $T$ lies on $\overline{FG}$. It is well-known (by homothety) that $I$ lies on $\overline{DM}$. Then, by incenter-excenter, $N$ lies on $(IBCI_A)$, which has diameter $\overline{II_A}$. Let the radical axis of $(IBCI_A)$ and the $A$-excircle be $\ell$, which is also the polar of $I$ with respect to the excircle. Since $I$ lies on the polar of $T$, $T$ lies on $\ell$. By radical center on $(ABC)$, $(IBCI_A)$, and the $A$-excircle, $\overline{BC}$, $\overline{FG}$, and $\ell$ concur, hence $T$ lies on $\overline{FG}$ as well; done. $\blacksquare$ Remark: During the solve process I went through every single ARCH hint (not at the same time) only to discover, on each of them, that I had already done what the hint told me to :thonk:

Let $M$ be the midpoint of $AE$, and let $\omega$ denote the $A$-excircle. Let $X = MD \cap \omega$, and define $V$ and $N$ as the intersections of $DF$ and $DG$ with the perpendicular bisector of $AE$. Let the tangents to $\omega$ from $D$ and $X$ meet at $T$. Projecting through $D$ it suffices to show, $$-1 = (VN,MP_\infty) \overset{D}{=} (FG,XD)$$or equivalently that $GXGD$ is harmonic. It then suffices to show $T-F-G$. Before we begin we provide a well-known lemma: $MD$ passes through $I$, the incenter of $ABC$. Now letting $L$ be the midpoint of $DX$, we find that $\angle KLM = 90$, where $K$ is the $A$-excenter. Then if $Y, Z = (IBC) \cap \omega$, we find by radical center that $BC$, $GF$ and $YZ$ concur. Then it would suffice to show that $T \in YZ$. However note by La Hire's it suffices to show that $I$ lies on the polar of $T$, namely $DX$, which is clearly true.

If $EVAN$ is a rhombus $EV = AV, EN = VN$ imply that $V, N$ lie on the perpendicular bisector of $AE.$ Define $V$ and $N$ to be the intersections with $DG, DF.$ It suffices to prove that $AV = AN$, equivalently the midpoint of $M$ of $AE$ is the midpoint of $AV, AN.$ Lemma 1: Let $K, L$ be the other extouch points of the $A$-excircle with $\overline{AB}, \overline{AC}.$ Define $D'= \overline{KL} \cap \overline{BC}.$ Then $(B, C; D, D') = -1.$

Let $U$ be the intersection of the $A$-angle bisector with $BC.$ Since $\angle IBI_A = 90^\circ$ and $\angle ABI = \angle UBI,$ we have $(A, U; I, I_A) = -1.$ Now consider the polars, w.r.t the $A$-excircle, of these four points. Since they are collinear, their polars are concurrent to the pole of $\overline{AI}$ by LaHire. We can see that the polars form an angle harmonic bundle. So consider their projections onto $BC$; $\text{polar}(A) = \overline{KL}$ means $A \to D'$, $D \in \text{polar}(U)$ so $U \to D$, and $I_A \to \infty.$ Since the projection is a harmonic bundle, $I$ must map to the harmonic conjugate of $\infty$, which is $R$ the midpoint of $DD'.$ So $R \in \text{polar}(I)$ and by LaHire, $I \in \text{polar}(R).$ Define $H$ to be the other tangency of $R$ to the excircle. Then $I \in \overline{DH}.$ Lemma 2: $M, I, D$ are collinear.

Lemma 3: Let $R$ be the midpoint of $DD'$, $R$ lies on $\overline{GF}.$

Now we can finish; $MDH$ are collinear, $RD, RH$ are tangents to the excircle, and $R, G, F$ are collinear. So, $$(N, V; M, \infty) \stackrel{D}{=} (F, G; H, D) = -1 $$and $M$ is the midpoint of $N, V.$ $\blacksquare$

Let $\ell$ be the line through $I$ parallel to $\overline{BC}$. It suffices to show that if lines $FD$ and $GD$ intersect $\ell$ at $X$ and $Y$ respectively, the midpoint of $\overline{XY}$ is $I$. Let $P$ and $Q$ be the intersections of lines $FD$ and $GD$ with $(ABC)$, respectively. By Reim's theorem, $BCQP$ is an isosceles trapezoid. In fact, we will show, via length calculations, that line $PQ$ is exactly halfway between parallel lines $BC$ and $XY$. In particular, the distance from $Q$ to line $\overline{BC}$ is \begin{align*} & \sin ( \angle QDC ) \cdot QD \\ = ~ & \sin (\angle DFG) \cdot \frac{DC \cdot BD}{DG} \\ = ~ & \frac{DC \cdot BD}{\frac{DG}{\sin (\angle DFG)}} \\ = ~ & \frac{DC \cdot BD}{2r_A} \\ = ~ & \frac{r}{2},

Thus, a homothety of factor $2$ centered at $D$ sends $P$ to $X$ and $Q$ to $Y$; thus, the perpendicular bisector of $\overline{PQ}$ is sent to the perpendicular bisector of $\overline{XY}$. But the perpendicular bisector of $\overline{PQ}$ is the perpendicular bisector of $\overline{BC}$; a homothety of factor $2$ centered at $D$ sends the perpendicular bisector of $\overline{BC}$ to the line through $I$ perpendicular to $\ell$. In particular, $I$ is the midpoint of $\overline{XY}$ as desired.

We set $N$ and $V$ to be the intersections of the perpendicular bisector of $AE$ with $DF$ and $DG$. It remains to show \[-1 = (VN;L \infty) \overset{D}{=} (GF;KD),\] where $L$ is the midpoint of $AE$ and $K = LD \cap \omega_A$. In other words, we need $XK$ to be tangent to the excircle, where $X = BC \cap GF$. From Midpoint of the Altitude lemma, $I$ lies on $KL$. Suppose $(II_A)$ meets the excircle at two points $P$ and $Q$. Radical axis theorem on the excircle, the circumcircle, and $(II_A)$ tells us $PQ$ passes through $X$. Finally, noting that $IP$ and $IQ$ are tangents to the excircle, $KPDQ$ is harmonic, which implies the desired. $\blacksquare$

Got the exact same solution as EVAN on accident Let $M$ be the midpoint of $AE$. It is well-known that $M$, $I$, and $D$ are collinear. Let $MID$ intersect the $A$-excircle again at $P$ and let $(II_A)$ intersect the $A$-excircle at $X$ and $Y$. By radical axis, we know that $BC$, $XY$, $FG$ are concurrent. Call this concurrency point $K$. We claim that $N$ is the intersection of the perpendicular bisector of $AE$ with $DF$, and $V$ similarly with $DG$. Take all poles and polars wrt the $A$-excircle. Since $KD$ is tangent to the $A$-excircle, $D$ lies on the polar of $K$. Also, note that $\angle IXI_A = IYI_A = 90^\circ$, hence $I$ is the pole of $XYK$, and by La Hire's, this implies that $I$ is on the polar of $K$. Hence the polar of $K$ is $MIDP$ and so \[ -1 = (DP;FG) \stackrel D= (P_\infty M;NV) \]where $P_\infty$ is the point of infinity along line $BC$. Thus $M$ is the midpoint of $AE$, $NV$, and $AE \perp NV$, and so $EVAN$ is a rhombus.

[asy][asy] import olympiad; import cse5; defaultpen(fontsize(10pt)); usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp"); size(8cm); pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) { r=r/2; pair Bp=unit(B-A)*r+A; pair Cp=unit(C-A)*r+A; pair P=Bp+Cp-A; D(Bp--P--Cp,q); return A; } pointpen=black+linewidth(2); pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0); pen polyfill=polyline+opacity(0.1); pen angleline=linewidth(pathpen)+rgb(0,0.4,0); pen anglefill=angleline+opacity(0.4); markscalefactor=0.01; size(15cm); pair A=dir(120),B=dir(210),C=dir(330); // filldraw(A--B--C--cycle,polyfill,polyline); D(A--B--C--cycle,polyline); pair I=incenter(A,B,C); pair I_A=2*circumcenter(B,I,C)-I; path exc=CP(I_A,foot(I_A,B,C)); path cir=circumcircle(A,B,C); pair F=IP(exc,cir,1); pair G=IP(exc,cir,0); D(exc); D(cir); pair D=foot(I_A,B,C),E=foot(A,B,C); pair M=(A+E)/2; D(B--foot(I_A,A,B)); D(C--foot(I_A,A,C)); D(A--E); pair V=extension(D,G,M,M+B-E),N=extension(D,F,M,M+C-E); D(V--N); D(F--N); D(G--V); D(D--M); pair L=2*foot(I_A,M,D)-D; pair X=(D+L)/2; pair Y=extension(B,C,F,G); D(D--L); D(C--Y--L); D(Y--I_A); D(circumcircle(B,I,C)); D("A",D(A),A); D("B",D(B),W); D("C",D(C),dir(0)); D("I",D(I),W); D("I_A",D(I_A),S); D("F",D(F),dir(181)); D("G",D(G),dir(-80)); D("D",D(D),dir(-98)); D("E",D(E),S); D("M",D(M),dir(48)); D("V",D(V),W); D("N",D(N),dir(0)); D("L",D(L),dir(0)); D("X",D(X),dir(3)); D("Y",D(Y),dir(45)); [/asy][/asy] Let $V$ and $N$ be the intersections of $DG$ and $DF$ with the perpendicular bisector of $\overline{AE}$, let $L$ be the second intersection of line $MD$ with the $A$-excircle and let $M$ be the midpoint of $\overline{AE}$. It suffices to show $MV = MN$, or equivalently that $(VN; M\infty) = -1$. Note that it is well known that $M$, $I$ and $D$ are collinear. Let $X$ be the midpoint of $\overline{DL}$ and $Y$ be $DD \cap LL$. Note that $\angle IXI_A = 90^\circ$ and thus $X$ lies on $(BIC)$. Additionally, $I_A L \perp YL$ and $XL \perp YI_A$, so we have \[YL^2 = YX \cdot YI_A = YB \cdot YC = YF \cdot YG\text{.}\]Thus $F$, $G$ and $Y$ are collinear and so $(DL; FG) = -1$. Projecting at $D$ gives us the desired result.

Let $M$ be the midpoint of $AE$. Then our condition translates $M$ being the midpoint of $V$ and $N$, where $V$ and $N$ are the intersections of the perpendicular bisector of $AE$ with $DG$ and $DF$. Note that by EGMO $4.12$ we have $M$, $D$, and the incenter $I$ collinear. Let $Y$ be the second intersection of $MD$ with the excircle. Since we have $(V, N; M, P_{\infty}) \overset{D}= (G, F; Y, D)$. We wish to show that $GFYD$ is a harmonic quadrilateral, or that the tangents at $F$ and $G$ wrt the excircle concur with $YD$. Let circle $(II_A)$ intersect the excircle at $P$ and $Q$. Then by Fact $5$, $(II_A)$ passes through $B$ and $C$. Notice that the pole of $XY$ is $I$, which implies that the polar of $R$ passes through $I$(by La Hire's), which implies that $GYFD$ is harmonic, so we are done.

Diagram has points $V$ and $N$ omitted.

in the finish before.

oops Let $\ell$ be the perpendicular bisector of $AE,$ and let $DG$ and $DF$ intersect $\ell$ at points $V$ and $N,$ respectively. If $M$ is the midpoint of $AE,$ we will also show that $M$ is the midpoint of $VN,$ which will evidently show that $EVAN$ is a rhombus. Let $I$ be the incenter; it is well-known that $M,I,D$ are collinear. Let the line $\overline{MID}$ intersect the $A$-excircle $\omega$ again at point $P$. Claim: (External version of ISL 2002 G7) $(BCP)$ is tangent to $\omega$. Proof: Let $(BCP)$ intersect $ID$ again at point $Z.$ We first show that $Z$ is the midpoint of $ID.$ Indeed, if $W$ is the midpoint of $DP,$ then Power of a Point implies $$DB \cdot DC = DZ \cdot DP = DI \cdot DW$$since $BICWI_A$ is cyclic ($\angle IWI_A = 90^\circ$). Since $DP = 2DW,$ we see that $DI = 2DZ,$ implying that $Z$ is the midpoint of $DI.$ Then the Shooting Lemma implies the result. Now, let the tangent at $P$ to $\omega$ intersect $BC$ again at point $X.$ Radical center on $(ABC), \omega, (BCP)$ implies that $F,G,X$ are collinear. Noting that $XD$ is tangent to $\omega$ as well, we discover that $DFPG$ is harmonic. Then $$(F,G;P,D) \stackrel{D}{=} (N,V;M,\infty) = -1,$$and we are done.

Instead of proving existance, lets try to construct a rhombus. Specifically, let $M$ be midpoint of $AE,$ and perpendicular bisector of $AE$ intersect $DF$ and $DG$ at $N$ and $V.$ We would like to prove $M$ is midpoint of $VN.$ Let the intersection between $DM$ and the $A$ excircle be $P.$ We want to prove $$-1=(V, N; M, P_{\infty})\overset{D}{=}(G, F; P, D).$$Let tangent at $DD$ and $FG$ intersect at $X.$ We want to show $DP$ or $DI$ (points $P, D, I, M$ are collinear follows from homothethy at $D$) is the polar of $X.$ Since $X \in DD,$ the polar of $D$ contains $X.$ Let's now consider the polar of $I.$ The polar is the line between the intersection between the circle with diameter $II_A$ and the excircle; let these intersections be $Y,Z$. We notice that by the incenter-excenter lemma, the midpoint of $II_A$ lies on $(ABC)$, and also $BCII_A$ is cyclic. Thus, by Radical Axis Theorem, $BC, FG, YZ$ concur, at $X$! Thus, the polar of $I$ contains $X.$ Hence, the polar of $X$ is $DI$, and we are done.

First of all, redefine $V,N$ as the intersections of the perpendicular bisector of $\overline{AE}$ with $\overline{DF}$, $\overline{DG}$. Denote the midpoint of $\overline{AE}$, incenter, and $A$-excenter as $M$, $I$, and $I_A$, respectively. Let $\Omega$, $\omega$, and $\omega_A$ be the circumcenter, $(II_A)$, and $A$-excircle, respectively. Finally, suppose that $\omega$ intersects $\omega_A$ at points $P$ and $Q$, that $\overline{MD}$ intersects $\omega_A$ at a second point $K$, and that the radical center of $\Omega$, $\omega$, and $\omega_A$ is $X$. We wish to show that $M$ is the midpoint of $\overline{VN}$, which is equivalent to \[-1 = (V,N;M,\infty) \overset{D}{=} (F,G;K,D).\] In other words, we wish to show that $DFKG$ is harmonic. Hence, it suffices to show that $\overline{XK}$ is tangent to $\omega_A$. Claim: $DPKQ$ is harmonic Proof: Recall that $I$ lies on $\overline{MD}$. Then, note that $\angle IPI_A = \angle IQI_A = 90^\circ$, which means $IP$ and $IQ$ are tangent to $\omega_A$. Hence, we are done. $\square$ Since $DPKQ$ is harmonic, we know that $\overline{DD}$, $\overline{PQ}$, $\overline{KK}$ concur, and we also know that the concurrency point is $X$. Therefore, $\overline{XK}$ is tangent to $\omega_A$ as desired. $\square$

Solved with SomeonesPenguin Denote by $\omega_A$ the A-excircle, let $I$ be the incenter and let $U$ and $V$ be the intersection of circle $(BIC)$ with $\omega_A$. Let $S$ be the midpoint of $AE$, $FG\cap BC=\{J\}$ and let $T$ be on $\omega_A$ such that $JT$ it tangent to $\omega_A$. Claim 1. $S$, $I$, $D$ and $T$ are collinear. Proof: Note that from homothety $D$, $I$ and $S$ are collinear. Note that $IU$ and $IV$ are tangent to $\omega_A$ and $J$ is the radical center of circles $\omega_A$, $(ABC)$ and $(BIC)$, so it lies on $UV$. Since $JT$ and $JD$ are tangent to $\omega_A$, it follows that $UDVT$ is harmonic, hence $I$ lies on $DT$. $\square$ Finally, the conclusion follows by projecting the harmonic quadrilateral $FDGT$ through $D$ on the line parallel to $BC$ passing through $S$.$\blacksquare$