sol with PoP and a lot of reflections: [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=mediumblue; pen sec=purple; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=paleblue; pair O, Ia, Ib, Ic, A, B, C, I, D, X, W, Y, W1, Y1, M, N, U, V, X1, X3, P, Q; O=(0, 0); Ia=dir(105); Ib=dir(210); Ic=dir(330); A=foot(Ia,Ib,Ic); B=foot(Ib,Ia,Ic); C=foot(Ic,Ia,Ib); I=intersectionpoint(A--Ia, B--Ib); X=C+0.25(C-B); X1=foot(X,A,Ia); Y=2*X1-X; X3=foot(X,Ib,Ic); W=2*X3-X; D=intersectionpoint(Y--(Y+100*(Y-A)), circumcircle(A, B, C)); P=intersectionpoint(X--Y, circumcircle(D, I, Ia)); Q=intersectionpoint(X--W, circumcircle(D, Ib, Ic)); M=(I+Ia)/2; N=(Ib+Ic)/2; Y1=2M-X; W1=2N-X; U=intersectionpoint(X--M, circumcircle(D,I,Ia)); V=intersectionpoint(X--N, circumcircle(D,Ib,Ic)); draw(Ia--Ib--Ic--cycle,pri); draw(A--Ia, pri); draw(B--Ib, pri); draw(C--Ic, pri); draw(A--B--C--cycle, pri); draw(circumcircle(A, B, C), pri); draw(C--X, pri); draw(circumcircle(D, I, Ia), tri); draw(circumcircle(D, Ib, Ic), tri); draw(X--Y, dashed+sec); draw(X--W, dashed+sec); draw(X--Y1,dashed+sec); draw(X--W1,dashed+sec); draw(D--W,pri); draw(A--X,pri); label("$I_A$",Ia,dir(90)); label("$I_B$",Ib,dir(180)); label("$I_C$",Ic,dir(0)); label("$A$",A,dir(270)); label("$B$",B,dir(60)); label("$C$",C,dir(120)); label("$I$",I,dir(240)); label("$D$",D,dir(0)); label("$X$",X,dir(150)); label("$W$",W,dir(240)); label("$Y$",Y,dir(300)); label("$W'$",W1,dir(300)); label("$Y'$",Y1,dir(60)); label("$M$",M,dir(120)); label("$N$",N,dir(270)); label("$U$",U,dir(180)); label("$V$",V,dir(270)); label("$P$",P,dir(60)); label("$Q$",Q,dir(180)); [/asy][/asy] Let $X$ be the point on $BC$ such that $\angle{CAX} = \angle{BAD}$; then it suffices to show that $X = E$, or that $X$ has equal power wrt $(DII_A), (DI_BI_C)$. Note that $\triangle{ABC}$ is the orthic triangle of $\triangle{I_AI_BI_C}$, and $\triangle{ADB} \sim \triangle{ACX}$, $\triangle{AIC} \sim \triangle{ABI_A}$, $\triangle{ACI_B} \sim \triangle{AI_CB}$ (latter two are classic properties of the orthic config) so $AD \cdot AX = AB \cdot AC = AI \cdot AI_A = AI_B \cdot AI_C$. Thus, if $Y, W$ are the reflections of $X$ over $II_A, I_BI_C$ respectively, $Y \in (DII_A), W \in (DI_BI_C)$. If $Y', W' =$ reflections of $X$ over midpoints of $II_A, I_BI_C$ then $IYY'I_A, I_BWW'I_C$ are cyclic isosceles trapezoids so if $P, Q = XY \cap (DII_A), XW \cap (DI_BI_C)$ then $PY', QW'$ are diameters of the two circles. Furthermore, $\angle{DPY'} = 90^{\circ} - \angle{DY'P} = 90^{\circ} - \angle{PYA} = \angle{DAI}$. By the nine point circle, the midpoints of $II_A, I_BI_C$ are on $(ABC)$, and so $U = XY' \cap (ABC)$ is on $(DII_A)$; similarly, $V = XW' \cap (ABC)$ is on $(DI_BI_C)$. Thus, $Pow_{(DII_A)}(X) = 2 \cdot Pow_{(ABC)}(X) = Pow_{(DI_BI_C)}(X)$, and we are done. $\square$

Let $O_1$ and $O_2$ be the circumcenters of $DII_A$ and $DI_BI_C$. Let $X$ be the point on $BC$ satisfying $\angle BAD=\angle XAC$. We will use Cartesian coordinates. Let $A=(0,0)$, $I=(0,-4ab)$, $I_A=(0,4cd)$, $I_B=(4ad,0)$, $I_C=(4bc,0)$. The circumcenter of $ABC$ is the circumcenter of $(0,0)$, $(0,2cd-2ab)$, and $(2ad+2bc,0)$, which is $(ad+bc,cd-ab)$. If $D=(x,y)$ then $(x-ad-bc)^2+(y-cd+ab)^2=(ad+bc)^2+(ab-cd)^2$, or $$x^2+y^2=2x(ad+bc)+2y(cd-ab)=2adx+2bcx+2cdy-2aby.$$ Now, compute $$O_1=\left(\frac{x^2+(y+4ab)(y-4cd)}{2x},2cd-2ab\right),$$ $$O_2=\left(2ad+2bc,\frac{y^2+(x-4ad)(x-4bc)}{2y}\right),$$ and $$X=\left(\frac{-16abcdx}{x^2+y^2},\frac{16abcdy}{x^2+y^2}\right).$$ The slope of $DX$ is $$\frac{y(x^2+y^2+16abcd)}{x(x^2+y^2-16abcd)}.$$ The slope of $O_1O_2$ is $$\frac{x(x^2+y^2+16abcd-4adx-4bcx-4cdy+4aby)}{y(-x^2-y^2+16abcd+4adx+4bcx+4cdy-4aby)}=\frac{x(-x^2-y^2+16abcd)}{y(x^2+y^2+16abcd)}.$$ The product of the two slopes is $-1$, so $DX\perp O_1O_2$. Therefore, $X$ lies on $DF$, so $E=X$, which means $\angle BAD=\angle EAC$.

We use barycentric coordinates. Let $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Then $I=(a,b,c)$, $I_a=(-a,b,c)$, and similarly for the others. Let $D=(p,q,r)$. Then $-a^2qr-b^2pr-c^2pq=0$. If the equation of $(DII_A)$ is $-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+wy+vz)=0$, we get that $ua+wb+vc=abc$ and $-ua+wb+vc=-abc$, so $u=bc$ and $wb+vc=0$. We have that $bcp+wq+vr=0$ as well, so we can solve the system and we have $w=\frac{bc^2p}{cq-rb}$ and $v=\frac{-b^2cp}{cq-rb}$. We do the same with $(DI_BI_C)$. Let the equation be $-a^2yz-b^2xz-c^2xy+(x+y+z)(dx+ey+fz)=0$, and we get that $-abc=da-(eb-fc)=da+(eb-fc)$, so $d=-bc$ and $eb=fc$. We also have $dp+eq+fr=0$, so $-bcp+eq+fr=0$. Thus $e=\frac{bc^2p}{cq+rb}$ and $f=\frac{b^2cp}{cq+rb}$. Note that the radical axis of the two circles is $DF$, which is $(d-u)x+(e-w)y+(f-v)z=0$ and we need to find this intersected with $BC$ which is $x=0$. Thus $E=(0,f-v,e-w)$. To prove the desired condition, we need $D*$, the isogonal conjugate of $D$, $E$, and $A$ to be collinear. We have $D*=\left(\frac{a^2}{p},\frac{b^2}{q},\frac{c^2}{r}\right)$. Thus we need to prove $E=\left(0,\frac{b^2}{q},\frac{c^2}{r}\right)$. This is trivial from our previous computations and a little work.

what the late USAMO 2023/6, orthic version wrote: Suppose $\triangle ABC{}$ is a triangle with orthic triangle $DEF$ and orthocenter $H{}$. Point $P$ lies on the nine-point circle, and point $Q$ lies on $EF$ such that $DP$ and $DQ$ are isogonal wrt $\angle EDF$. Show that $Q$ lies on the radical axis of $(PAH)$ and $(PBC)$. oops diagram later Redefine $Q$ as the intersection of said radical axis with $EF$. We will show that $DP$ and $DQ$ are isogonal wrt $\angle EDF$. Let $R = AH \cap EF$, and let $S = BC \cap EF$. In addition, let $\ell_R$ be the radical axis of $(DEF)$ and $(PAH)$, and let $\ell_S$ be the radical axis of $(DEF)$ and $(PBC)$. Note that $$\operatorname{Pow}_{(PAH)} (R) = RA \cdot RH = RE \cdot RF = \operatorname{Pow}_{(DEF)} (R),$$ where lengths are directed and the second equality comes from Power of a Point on $(AEHF)$. Thus $R \in \ell_R$. Similarly, $S \in \ell_S$. Now let $\ell_S$ hit $(PAH)$ again at $X$, and let $\ell_R$ hit $(PBC)$ again at $Y$. We claim that $PDXY$ is cyclic. To see this, consider the circle $\omega$ such that $$\operatorname{Pow}_\omega(K) = \operatorname{Pow}_{(PBC)} (K) + \operatorname{Pow}_{(PAH)} (K) - \operatorname{Pow}_{(DEF)} (K),$$ for all points $K$ (this exists by say, Cartesian coordinates). We can check that $\omega$ passes through $P$ since each power is $0$, and $\omega$ passes through $D$ since $$\operatorname{Pow}_{(PAH)} (D) + \operatorname{Pow}_{(PBC)} (D) = DA \cdot DH + DB \cdot DC = 0.$$ In addition, $\omega$ passes through $X$ since $X$ lies on $(PAH)$ and the radical axis of $(PBC)$ and $(DEF)$. Similarly, it passes through $Y$, so $\omega$ is the desired circle. To finish, note that by Radical Axis Theorem, the six pairwise radical axes of the circles $(PAH)$, $(PBC)$, $(DEF)$, and $(PDXY)$, form a complete quadrangle with vertices at their triple-wise radical centers. By Desargues's Involution Theorem on this complete quadrangle and line $EF$, there is an involution on $EF$ swapping $Q \leftrightarrow PD \cap EF$, $R \leftrightarrow R$, and $S \leftrightarrow S$. Projecting onto the pencil of lines through $D$, there is an involution on said pencil swapping $DQ \leftrightarrow DP$ and fixing both $DA$ and $DB$. The latter two imply it is reflection over line $DA$, so we are done. $\blacksquare$

Probably an Ankan problem tbh

barybash We use barycentric coordinates on $\triangle{ABC}$. Let $D=(\alpha:\beta:\gamma)$. Then since $D$ lies on the circumcircle of $\triangle{ABC}$, we have that $a^2\beta\gamma+b^2\alpha\gamma+c^2\alpha\beta=0$. Let $E'$ be the intersection of the reflection of $AD$ over $AI$ with $BC$. Then $E'=(0:b^2/\beta:c^2/\gamma)$. We wish to show that $E'$ lies on the radical axis of the circumcircles $\omega_1,\omega_2$ of $\triangle{DII_a},\triangle{DI_bI_c}$ respectively, i.e. that $\operatorname{Pow}_{\omega_1}(E')=\operatorname{Pow}_{\omega_2}(E')$. Let the equation of $\omega_i$ be $a^2yz+b^2xz+c^2xy-(x+y+z)(u_ix+v_iy+w_iz)=0$. Then we get the equations $$\begin{align*} a^2bc-ab^2c-abc^2&=(-a+b+c)(-u_1a+v_1b+w_1c)\\ a^2bc+ab^2c+abc^2&=(a+b+c)(u_1a+v_1b+w_1c)\\ 0=a^2\beta\gamma+\alpha b^2\gamma +\alpha\beta c^2&=(\alpha+\beta+\gamma)(u_1\alpha+v_1\beta+w_1\gamma)\\ -a^2bc+ab^2c-abc^2&=(a-b+c)(u_2a-v_2b+w_2c)\\ -a^2bc-ab^2c+abc^2&=(a+b-c)(u_2a+v_2b-w_2c)\\ 0=a^2\beta\gamma+\alpha b^2\gamma +\alpha\beta c^2&=(\alpha+\beta+\gamma)(u_2\alpha+v_2\beta+w_2\gamma)\\ \end{align*}$$ since $I=(a:b:c),I_a=(-a:b:c),I_b=(a:-b:c),I_c=(a:b:-c)$. These simplify to $$\begin{align*} -abc&=-u_1a+v_1b+w_1c\\ abc&=u_1a+v_1b+w_1c\\ 0&=u_1\alpha+v_1\beta+w_1\gamma\\ -abc&=u_2a-v_2b+w_2c\\ -abc&=u_2a+v_2b-w_2c\\ 0&=u_2\alpha+v_2\beta+w_2\gamma\\ \end{align*}$$ Thus, we get that $u_1=bc,u_2=-bc,v_1b+w_1c=v_2b-w_2c=0$. Then solving for $v_1,v_2,w_1,w_2$ gives us $$\begin{align*} v_1=\frac{\alpha bc^2}{b\gamma-c\beta}&,\qquad w_1=-\frac{\alpha b^2c}{b\gamma-c\beta}\\ v_2=\frac{\alpha bc^2}{b\gamma+c\beta}&,\qquad w_2=\frac{\alpha b^2c}{b\gamma+c\beta}\\ \end{align*}$$ Finally, we have that $$\begin{align*} u_1\cdot 0+v_1\cdot \frac{b^2}{\beta}+w_1\cdot\frac{c^2}{\gamma}&=\frac{\alpha bc^2}{b\gamma-c\beta}\cdot \frac{b^2}{\beta}-\frac{\alpha b^2c}{b\gamma-c\beta}\cdot\frac{c^2}{\gamma}\\ &=\frac{\frac{\alpha b^3c^2\gamma-\alpha b^2c^3\beta}{\beta\gamma}}{b\gamma-c\beta}\\ &=\frac{\frac{\alpha b^2c^2(b\gamma-c\beta)}{\beta\gamma}}{b\gamma-c\beta}\\ &=\frac{\alpha b^2c^2}{\beta\gamma}\\ \end{align*}$$ Similarly, we can compute $$\begin{align*} u_2\cdot 0+v_2\cdot \frac{b^2}{\beta}+w_2\cdot\frac{c^2}{\gamma}&=\frac{\alpha bc^2}{b\gamma+c\beta}\cdot \frac{b^2}{\beta}+\frac{\alpha b^2c}{b\gamma+c\beta}\cdot\frac{c^2}{\gamma}\\ &=\frac{\frac{\alpha b^3c^2\gamma+\alpha b^2c^3\beta}{\beta\gamma}}{b\gamma+c\beta}\\ &=\frac{\frac{\alpha b^2c^2(b\gamma+c\beta)}{\beta\gamma}}{b\gamma+c\beta}\\ &=\frac{\alpha b^2c^2}{\beta\gamma}\\ \end{align*}$$ Thus, since these are equal, $E'$ lies on the radical axis of $\omega_1$ and $\omega_2$. However, this is the line $DF$, so $E'=DF\cap BC=E$. Thus, using directed angles we get that $$\measuredangle BAD=\measuredangle BAI+\measuredangle IAD=\measuredangle IAC+\measuredangle E'AI=\measuredangle E'AC=\measuredangle EAC$$ so $\angle BAD=\angle EAC$. $_\blacksquare$ Note: by the assumptions on the location of $D$, none of $\beta,\gamma,b\gamma-c\beta,b\gamma+c\beta$ are zero.

Define $E$ as the point on $BC$ such that $DAB = EAC$. Then, $ADB = ACB = ACE$, so $ADB$ is similar to $ACE$. This means that $$AD\cdot AE = AB\cdot AC = AI\cdot AI_A = AI_B\cdot AI_C$$ Thus, since $AD$ and $AE$ are isogonal, reflecting $E$ over $II_A$ and $I_BI_C$ to $E'$ and $E''$ give points on $AD$. Power of a Point through A shows that $E'DII_A$ and $E''DI_BI_C$ are cyclic. Additionally, it is well known that $II_ABC$ and $I_BI_CBC$ are cyclic. Now, we will apply Linearity of Power of a Point on lines $EE'$ and $EE''$. Since the midpoint of $EE'$ lies on $II_A$, the radical axis of these two circles, we get $$Pow_{(DII_A)}(E) - Pow_{(BIC)}(E) + Pow_{(DII_A)}(E') - Pow_{(BIC)}(E') = 0$$ However, since $(BIC)$ is the circle with diameter $II_A$, the power with respect to this circle is the same for $E$ and $E'$. Additionally, $Pow(DII_A)(E') = 0$. Thus, we get $$Pow_{(DII_A)}(E) = 2Pow_{(BIC)}(E) = 2\cdot EB \cdot EC$$ If we do the same thing for $EE''$, we get $$Pow_{(DI_BI_C)}(E) - Pow_{(BI_BC)}(E) + Pow_{(DI_BI_C)}(E'') - Pow_{(BI_BC)}(E') = 0$$ which for the same reason, gives $$Pow_{(DI_BI_C)}(E) = 2Pow_{(BI_BC)}(E) = 2\cdot EB \cdot EC$$ Thus, the power of $E$ is the same with respect to both circles, so $E$ is on their radical axis, namely $DF$.

'cause every night i lie in bed and coordinate bashing fills my head...

sort of killed by moving points ? (also in-contest sol lol)

v4913 wrote: Let $ABC$ be a triangle with incenter $I$ and excenters $I_a, I_b,$, and $I_c$ opposite $A, B$, and $C$, respectively. Let $D$ be an arbitrary point on the circumcircle of $\triangle{ABC}$ that does not lie on any of the lines $II_a, I_bI_c$, or $BC$. Suppose the circumcircles of $\triangle{DII_a}$ and $\triangle{DI_bI_c}$ intersect at two distinct points $D$ and $F$. If $E$ is the intersection of lines $DF$ and $BC$, prove that $\angle{BAD} = \angle{EAC}$. This was my fav problem of the whole test.

cinnamon_e wrote: he didn't even make aime so?

so he's trolling

Make a way, projective balistic missile incoming!. (Generalize for all $D$) Fix $\triangle ABC$ and make $D$ move in $(ABC)$ now let $E'$ be the inverse of $D$ in $\sqrt{bc}$ inversion and let $O_1,O_2$ the centers of $(DI_bI_c)$ and $(DII_a)$ respectivily. Now note that $\text{deg} E'=1$, $\text{deg} O_2=2$ (becuase it moves in the straight bisector of $II_a$ without coming back but covers it twice), $\text{deg} O_1=2$ (becuase in this case even if it lies in the bisector of $I_bI_c$ we notice that it goes back when $(I_bI_cD)$ is tangent to $(ABC)$ so we can say it lies on a degenerate parabola), hence this means $\text{deg} F=2 \text{max} \{\text{deg} O_1, \text{deg} O_2, \text{deg} D \}=4$ so if we want to prove $D,E',F$ colinear we need $4+2+1+1=8$ cases! Case 1: $D$ midpoint of arc $BAC$ Clearly $(DII_a) \cap I_bI_c=F \ne D$ which is enough since $E'$ goes to $I_bI_c \cap BC$ (use McLaurin to verify) Case 2,3: $D=B,C$ (wlog $D=B$) Then $F=C$ but also $E'=C$ so we are done. Case 4: $D$ midpoint of minor arc $BC$ Then $F=II_a \cap (DI_bI_c)$ but $E'$ is $II_a \cap BC$ ($\sqrt{bc}$ inversion) so done. Case 5,6,7,8: $D$ is one of the intersections of either $(I_bII_a)$ or $(I_cII_a)$ with $(ABC)$ (wlog the one in minor arc $BC$ of $(I_bII_a)$) These are similar since the idea (notice $F$ is $I_b$ here) so let $E_1$ the inverse of $D$ w.r.t. $\sqrt{I_bI \cdot I_bB}$ inversion with center $I_b$ so now by PoP $I_aAE_1D$ is cyclic and also $B,E_1,C$ colinear by the inversion so by $\sqrt{bc}$ inversion we get $E_1=E'$ so we are done. Since we cleared all the cases it holds always thus we are done

geometry is keeping me awake

coordinate bashing fills my head

bary moment

Here is the harder version of this problem. I will post the solution to the harder version later. Harder than USAMO 2023 P6 wrote: Prove that the midpoint of $EF$ lies on $\odot(ABC)$. Solution to the original USAMO. Redefine $E$ to be the point on $BC$ such that $AI$ bisects $\angle DAE$. Let $O_1, O_2$ be the centers of $\odot(II_AD)$ and $\odot(I_BI_CD)$. Let $M$ and $N$ be the midpoints of $I_BI_C$ and $II_A$, respectively. [asy][asy] size(11cm,0); defaultpen(fontsize(10pt)); pair I_A = dir(115); pair I_B = dir(200); pair I_C = dir(340); pair A = foot(I_A, I_B, I_C); pair B = foot(I_B, I_A, I_C); pair C = foot(I_C, I_A, I_B); pair I = orthocenter(I_A, I_B, I_C); pair D = (I + dir(50))/2; pair E = extension(reflect(A,I)*D, A, B, C); pair E1 = reflect(A,I)*E; pair E2 = reflect(I_B,I_C)*E; pair O1 = circumcenter(I_A,I,D); pair O2 = circumcenter(I_B,I_C,D); pair K1 = 2*foot(O1,E,E1) - E1; pair K2 = 2*foot(O2,E,E2) - E2; pair S = extension(O1, (E1+K1)/2, O2, (E2+K2)/2); pair T = 2*circumcenter(A,B,C) - A; pair N = (I+I_A)/2; pair M = (I_B+I_C)/2; draw(E1--E--E2, linewidth(1)); draw(E2--D); draw(O1--T--O2--S--cycle, deepgreen+linewidth(1.2)); draw(S--(E2+K2)/2, deepgreen); draw(O1--(I+I_A)/2, deepgreen); draw(E--C,blue+linewidth(0.4)); draw(A--I_A,blue+linewidth(0.4)); draw(circumcircle(I_A,I,D), red+linewidth(1)); draw(circumcircle(A,B,C),blue+linewidth(0.4)); draw(arc(circumcenter(I_B,I_C,D), circumradius(I_B,I_C,D), degrees(E2-O2)+10, 540-degrees(E2-O2)-10, CW), red+linewidth(1)); draw(I_A--I_B--I_C--cycle, gray); draw(A--B--C--cycle, blue+linewidth(0.4)); dot("$A$",A,dir(270),blue); dot("$B$",B,NE,blue); dot("$C$",C,NW,blue); dot("$I$",I,1.7*dir(-70),blue); dot("$I_A$",I_A,dir(90)); dot("$I_B$",I_B,dir(180)); dot("$I_C$",I_C,dir(0)); dot("$D$",D,1.5*dir(30),red); dot("$E$",E,NW); dot("$E_1$",E1,NW); dot("$E_2$",E2,dir(180)); dot("$K_1$",K1,dir(-95),gray); dot("$K_2$",K2,dir(180),gray); dot("$O_1$",O1,dir(90),deepgreen); dot("$O_2$",O2,dir(-90),deepgreen); dot("$T$",T,dir(45),deepgreen); dot("$S$",S,dir(-90),deepgreen); dot("$M$",M,NW,blue); dot("$N$",N,dir(120),blue); [/asy][/asy] Claim. Let $E_1$ and $E_2$ be the reflections of $E$ across $II_A$ and $I_BI_C$, respectively. Then, $A, D, E_1, E_2$ are colinear. $E_1\in\odot(II_AD)$ and similarly $E_2\in\odot(I_BI_CD)$. Proof. The first part is obviously true because $AI$ (and thus $I_BI_C$) bisects $\angle DAE$. For the second part, note that since $\triangle ACE\sim\triangle ADB$, we have $AD\cdot AE = AB\cdot AC$, so $$AE_1\cdot AD = AE\cdot AD = AB\cdot AC = AI\cdot AI_A,$$ implying that $E_1\in\odot(DII_A)$. Similarly, $$AE_2\cdot AD = AE\cdot AD = AB\cdot AC = AI_B\cdot AI_C,$$ implying that $E_2\in\odot(DI_BI_C)$. $\blacksquare$ Claim. The circumcenter of $\triangle ADF$ is the midpoint of $O_1O_2$. Proof. We have $$\frac{\operatorname{Pow}(A, \odot(II_AD))}{\operatorname{Pow}(A, \odot(I_BI_CD))} = \frac{AI\cdot AI_A}{-AI_B\cdot AI_C} = -1,$$ so the result follows from the coaxial circle lemma. $\blacksquare$ Back to the main problem We let $EE_1$ intersects $\odot(II_AD)$ again at $K_1$, and let $EE_2$ intersects $\odot(I_BI_CD)$ again at $K_2$. It suffices to show that $E_1, E_2, K_1, K_2$ are concyclic. We do so by the following: let $S$ be the intersection of perpendicular bisectors of $E_1K_1$ and $E_2K_2$, then it suffices to show that $SE_1=SE_2$. Since $A$ is the midpoint of $E_1E_2$, this is equivalent to $\angle SAD=90^\circ$. To that end, we let $T = \overline{O_1N}\cap\overline{O_2M}$, which is the antipode of $A$ w.r.t. $\odot(ABC)$. Notice that $O_1TO_2S$ is a rectangle. Now, observe that Since $T$ and $A$ are antipodes, we have $TD\perp AD$. The midpoint $K$ of $ST$ is the midpoint of $O_1O_2$. From the second claim, the projection of $K$ onto $AD$ is the midpoint of $AD$. From these two items, we conclude that $SA\perp AD$, so we are done.

Rephrase the problem in terms of triangle $I_AI_BI_C$ (I've changed the point labellings, too): Reworded problem wrote: Let $ABC$ be a triangle and $DEF$ be its orthic triangle. Let $P \in (DEF)$ and $Q$ be on $FE$ such that $\angle QDF=\angle PDE$. Then prove that $Q$ belongs to the radical axis of $(BPC)$ and $(PAH)$. Let $X,Y$ be the projections of $Q$ on $BC,AH$, and $PD$ intersect $(BPC)$ and $(APH)$ at $Q'$ and $Q''$ respectively. Claim 1: $Q'$ is the reflection of $Q$ in $BC$ and $Q''$ is the reflection of $Q$ in $AH$. Proof: Note that since $DP,DQ$ are isogonal, triangles $FQD$ and $PDE$ are similar, and so $DP \cdot DQ=DF \cdot DE=DB \cdot DC=DP \cdot DQ',$ and so $Q'$ is the reflection of $Q$ in $BC$. Similarly $Q''$ is the reflection of $Q$ in $AH$ $\blacksquare$ To the problem, let $QX$ intersect $(BPC)$ again at $U \neq Q'$, and note that ${\rm pow} (Q,(BPC))=QQ' \cdot QU=2QX \cdot (UX-QX)=2QX \cdot UX-2QX^2=2(BX \cdot XC-QX^2),$ and similarly ${\rm pow} (Q,(PAH))=2(AY \cdot YH-QY^2),$ and so we are left to prove that $BX \cdot XC-QX^2=AY \cdot YH-QY^2$. Note that $BX \cdot XC+QY^2=(BD-QY)(CD+QY)+QY^2=(BD-CD)QY+BD \cdot CD$ and $AY \cdot YH+QX^2=(AD-QX)(QX-HD)+QX^2=(AD+HD)QX-AD \cdot HD,$ and so we are left to prove that $(BD-CD)QY+BD \cdot CD=(AD+HD)QX-AD \cdot HD$. Let $AD$ and $FE$ intersect at $R$. Note that $\angle QRY=90^\circ-\angle B+\angle C$. Therefore, $QY=RY\cot(\angle B-\angle C)=(RD-QX)\cot(\angle B-\angle C),$ and so we need to prove that (we also use that $BD \cdot CD=AD \cdot HD$) $(BD-DC)\cot(\angle B-\angle C)RD+2BD \cdot DC+((DC-DB)\cot(\angle B-\angle C)-(AD+DH))QX=0$ To make our live easier, we prove the following Claim: Claim 2: $(BD-DC)\cot(\angle C-\angle B)=AD+DH$. Proof: Note that $\dfrac{BD-DC}{AD+DH}=\dfrac{1}{\dfrac{AD}{DB}+\dfrac{DH}{DB}}-\dfrac{1}{\dfrac{AD}{DC}+\dfrac{DH}{DC}}=\dfrac{1}{\tan \angle B+\dfrac{1}{\tan \angle C}}-\dfrac{1}{\tan \angle C+\dfrac{1}{\tan \angle B}}=\dfrac{\tan \angle C-\tan \angle B}{\tan \angle B \tan \angle C+1}=\tan(\angle C-\angle B),$ as desired $\blacksquare$ Hence, the second summand of the LHS of the equality we have to prove is zero, and in order to prove that the first one is zero, too, we only need to prove that $(AD+DH)RD=2BD \cdot DC,$ or equivalently that $(AD+DH)RD=2DF \cdot DE$. However, note that $\dfrac{AD+DH}{DF}=\dfrac{\sin \angle C}{\cos \angle B}+\dfrac{\cos \angle C}{\sin \angle B}=\dfrac{\cos(\angle B-\angle C)}{\sin \angle B \cos \angle B}$ and $\dfrac{2DE}{RD}=\dfrac{2\cos(\angle B-\angle C)}{\sin \angle 2B}=\dfrac{\cos(\angle B-\angle C)}{\sin \angle B \cos \angle B},$ and so the two expressions are equal, as desired. The proof is finished.

@above please provide a reference to understand the godforsaken things you've written thanks

oVlad wrote: @above please provide a reference to understand the godforsaken things you've written thanks Isogonal pivotal cubic can be found at https://mathworld.wolfram.com/PivotalIsogonalCubic.html. Group law on Cubics is AFAIK equivalent to Group law on Elliptical Curves, which has many resources. Circle Points are the points at infinity of $(1 : i : 0)$ and $(i : 1 : 0)$, but the gist is that they are isogonal conjugates and a conic is a circle iff it goes through both circle points.

Wow bary so strong. It suffices to show that $D$ and $E$ are isogonal in $\angle BAC$. Let $\omega_1$ and $\omega_2$ denote $(DII_A)$ and $(DI_BI_C)$ respectively. Now we use barycentric coordinates. Take $\triangle A BC$ as reference. Then we must have $I = (a:b:c)$ and similarly $I_A = (-a:b:c)$, $I_B = (a:-b:c)$ and $I_C = (a:b:-c)$. Take $D = (x_1:y_1:z_1)$ satisfying, $$a^2y_1z_1 + b^2x_1z_1 + c^2x_1y_1 = 0.$$ Also take $E' = (0 : b^2z_1 : c^2y_1)$, which is isogonal to $D$ in $\angle BAC$. Then it suffices to show $$Pow_{\omega_1}(E') = Pow_{\omega_2}(E').$$ Now we will find the equation of $\omega_1$. Plugging points into the arbitrary circle equation we have, $$\begin{align*} -abc(a + b + c) + (a+b+c)(u_1a + v_1b + w_1c) &= 0\\ abc(-a + b + c) + (-a + b + c)(-u_1a + v_1b + w_1c) &= 0\\ -a^2y_1z_1 - b^2x_1z_1 - c^2x_1y_1 + (x_1+y_1+z_1)(u_1x_1 + v_1y_1 + w_1z_1) &= 0\\ \end{align*}$$ These relatively nicely simplify down to, $$\begin{align*} u_1a + v_1b + w_1c &= abc\\ -u_1a + v_1b + w_1c &= - abc\\ u_1x_1 + v_1y_1 + w_1z_1 &= 0 \end{align*}$$ Now subtracting the first two equations we find $u_1 = bc$. Then we have the new system, $$\begin{align*} v_1b + w_1c &= 0\\ bcx_1 + v_1y_1 + w_1z_1 &= 0 \end{align*}$$ The first equation implies we have $\frac{v_1}{w_1} = \frac{-c}{b} \implies w_1 = \frac{-b}{c}v_1$. Plugging this into the second we wish to find a solution to, $$\begin{align*} bcx_1 + v_1y_1 - \frac{b}{c}v_1z_1 &= 0\\ v_1\left(y_1 - \frac{b}{c}z_1 \right) &= -bcx_1\\ v_1 &= \frac{-bc^2x_1}{cy_1 - bz_1} \end{align*}$$ Then clearly we find $w_1 = \frac{b^2cx_1}{cy_1 - bz_1}$ so our circle has equation, $$\begin{align*} -a^2yz -b^2xz - c^2xy + (x+y+z)\left(bcx - \frac{bc^2x_1}{cy_1 - bz_1}y + \frac{b^2cx_1}{cy_1 - bz_1}z \right) &= 0 \end{align*}$$ Yay. Now we have to do the same thing for $\omega_2$. We get the system, $$\begin{align*} abc(a - b + c) + (a-b+c)(u_2a - v_2b + w_2c) &= 0\\ abc(a + b - c) + (a+b-c)(u_2a + v_2b - w_2c) &= 0\\ -a^2y_1z_1 - b^2x_1z_1 - c^2x_1y_1 + (x_1+y_1+z_1)(u_2x_1 + v_2y_1 + w_2z_1) &= 0\\ \end{align*}$$ This neatly simplifies to, $$\begin{align*} u_2a - v_2b + w_2c &= -abc\\ u_2a + v_2b - w_2c &= -abc\\ u_2x_1 + v_2y_1 + w_1z_1 &= 0 \end{align*}$$ Adding the first two equations give $u_2 = -bc$. Plugging this back in we have the system, $$\begin{align*} -v^2b + w^2c &= 0\\ -bcx_1 + v_2y_1 + w_1z_1 &= 0 \end{align*}$$ The first equation gives $\frac{v_2}{w_2} = \frac{c}{b} \implies w_2 = \frac{b}{c}v_2$. Plugging this into the second we have, $$\begin{align*} -bcx_1 + v_2y_1 + \frac{b}{c}v_2z_1 &= 0\\ v_2\left(y_1 + \frac{b}{c}z_1 \right) &= bcx_1\\ v_2 &= \frac{bc^2x_1}{cy_1 + bz_1} \end{align*}$$ Then we easily get $w_2 = \frac{b^2cx_1}{cy_1 + bz_1}$. Thus our circle has equation, $$-a^2yz - b^2xz - c^2xy + (x+y+z)\left(-bcx + \frac{bc^2x_1}{cy_1 + bz_1}y + \frac{b^2cx_1}{cy_1 + bz_1} \right) = 0$$ Now we're on the homestretch. Plugging in $E' = (0 :b^2z_1 : c^2y_1)$ into the circle equations we have, $$\begin{align*} Pow_{\omega_1}(E') &= -a^2b^2c^2z_1y_1 + (b^2z_1 + c^2y_1)\left(-\frac{b^3c^2x_1z_1}{cy_1-bz_1} + \frac{b^2c^3x_1y_1}{cy_1-bz_1} \right)\\ Pow_{\omega_2}(E') &= -a^2b^2c^2z_1y_1 + (b^2z_1 + c^2y_1) \left(\frac{b^3c^2x_1z_1}{cy_1 + bz_1} + \frac{b^2c^3x_1y_1}{cy_1 + bz_1} \right) \end{align*}$$ Thus it suffices to show, $$\begin{align*} \frac{b^2c^3x_1y_1 - b^3c^2x_1z_1}{cy_1 - bz_1} &= \frac{b^3c^2x_1z_1 + b^2c^3x_1y_1}{cy_1+bz_1}\\ \frac{b^2c^2x_1(cy_1 - bz_1)}{cy_1 - bz_1} &= \frac{b^2c^2x_1(bz_1 + cy_1)}{cy_1 + bz_1}\\ b^2c^2x_1 &= b^2c^2x_1 \end{align*}$$ so we are done. $\blacksquare$

Here is my solution (linpop): Re-define $E$ to be the point on $BC$ such that it is isogonal to $D$ with respect to $\angle BAC$. Thus, it suffice to prove $E$ lies on the radical axis of $(DII_A)$ and $(DI_BI_C)$. Let $E_1, E_2$ denote the reflections of $E$ with respect to $AI_A$ , $I_B I_C$, respectively. Note that (by Force-Overlaid Inversion): $$AI_A \cdot A I_B = AI \cdot AI_A = AE \cdot AD = A E_1 \cdot AD = A E_2 \cdot AD = AB \cdot AC$$ Hence, $DE_1 II_A$ and $DI_B E_2 I_C$ are cyclic. Now, define $f(X) = Pow(X, (D I_B I_C)) - Pow(X, (D I I_A))$. Thus, it suffice to prove that $f(E) = 0$. Now, we define: $$p(X) = Pow(X, (D I_B I_C)) - Pow(X, (I_B I_C BC))$$ $$g(X) = Pow(X, (I_B I_C BC)) - Pow(X, (IBI_AC))$$ $$h(X) = Pow(X, (IB I_A C)) - Pow(E, (D I I_A))$$ Note that $f = p+g+h$. Now, notice that: $p(E) = -p(E_2)$ as $E_2$ is the reflection of $E$ with respect to $I_B I_C$ which is the radical axis of $(I_B I_C BC)$ and $(DI_BI_C)$. Similarly, $h(E) = -h(E_1)$. Also, note that $g(E) = 0$ as $E$ lies on their radical axis. Hence; $$f(E) = p(E)+g(E)+h(E) = Pow(E_2, (I_B I_C BC)) - Pow (E_1, (IB I_A C))$$ $$= -Pow(E, (I_B I_C BC)) + Pow (E, (IB I_A C)) = 0$$

SMA665 wrote: I think this is the greatest USAMO no it is not, in my opinion USAMO 1990 was better and we win

let $AI_a$ intersect the circumcircle of $ABC$ at $F$, and intersect $BC$ at $G$. extend $AE$ to intersect the circumcircle of $ABC$ at $H$ we easily get that $ADC=ABC=ABE=AFC$, and by incenter and excenter properties, $BAG=FAC=BCF$ and $ACF~CGF~AGB$ we also have $CAD=CBD$, $BAH=BCH$, and $BF=CF=IF=I_aF$. we then prove $BDHC$ is an isoceles trapezoid via 2019 IMO P6 and get that $CHF$ and $BDF$ are congruent, and $CDF=BDF=CAH=BAD=CAE$

Redefine $E$ as the image of $D$ under force-overlaid inversion with respect to $\triangle ABC$. Let $f(\bullet) = \text{Pow}(\bullet, (DI_BI_C))-\text{Pow}(\bullet, (DII_A))$. We will show that $f(E)=0$. We write $f$ as a sum of linear functions $f_1$, $f_2$, and $f_3$ defined by $$f_1(\bullet) = \text{Pow}(\bullet, (DI_BI_C))-\text{Pow}(\bullet, (II_A))$$ $$f_2(\bullet) = \text{Pow}(\bullet, (II_A))-\text{Pow}(\bullet, (I_BI_C))$$ $$f_3(\bullet) = \text{Pow}(\bullet, (I_BI_C)) - \text{Pow}(\bullet, (DII_A)).$$ Note that $f_2$ vanishes at $E$, so it suffices to show that $g = f_1+f_3$ vanishes at $E$. Reflect $E$ over $AI_A$ to $E_1$, and $E$ over $I_BI_C$ to $E_2$. By force-overlaid inversion and power of a point, it is easy to see that $E_1$ lies on $(DII_A)$ and $E_2$ lies on $(DI_BI_C)$. Now note that $f_1(E) = -f_1(E_1)$ and $f_3(E)=-f_3(E_2)$ since the midpoints of $EE_1$ and $EE_2$ lie on the kernels of $f_1$ and $f_3$, respectively. We compute that $-f_1(E_1) = \text{Pow}(E_1, (II_A)) = \text{Pow}(E, (II_A))$ and $-f_3(E_2) = -\text{Pow}(E_2, (I_BI_C))=-\text{Pow}(E, (I_BI_C))$. Thus it suffices to show that $$\text{Pow}(E, (I_BI_C)) - \text{Pow}(E, (II_A)) = 0,$$ but this is clear as the radical axis of $(I_BI_C)$ and $(II_A)$ is $BC$, which $E$ lies on.

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actually zero mohs problem ! ! ! (half jokes???) Redefine $E$ as the point where $\angle BAD=\angle EAC$. We show that $E$ lies on the radical axis of the given circumcircles. Reflect $E$ across $A$-internal and external bisectors to $U$ and $V$. Notice that $U\in (DII_a)$ and $V\in (DI_bI_c)$. Let $R=EU\cap (DII_a)$ and $S=EV\cap (DI_bI_c)$. It suffices to show that $EU\cdot ER=EV\cdot ES$. Move $E$ linearly in $t$ on $BC$. Expressing in terms of degrees, we have: $$EU=1$$ $$ER=\frac{2}{1}$$ $$EV=1$$ $$ES=\frac{2}{1}$$ where we found degree for $ER$ (analogously, $ES$) by defining $M$ as midpoint of $EU$, then noticing that $MI\cdot MI_a=MU\cdot MR$. So now we just need to show something like $$\frac{3}{1}=\frac{3}{1}$$ which essentially means we need to prove a degree four polynomial is equal to zero. So it suffices to check five points for $E$. $E=B,C$ are easy. $E=AI\cap BC$ is easy. (Because $(DII_a)$ is a line.) $E=AI_bI_c\cap BC$ is easy. (Because $(DI_bI_c)$ is a line.) Notice that $\sqrt{bc}$ inversion means it's okay to prove that $(ADE)$, $(II_aE)$, and $(I_bI_cE)$ have another intersection point. Consider when $E\to \infty_{BC}$. Then the intersection point of $(I_bI_cE)$ and $(II_aE)$ approaches $A$ which is legit enough (bahahaha seriously though it should work?) to imply $E=\infty_{BC}$ is totally gaming. Done

synthetic progress $ABCH$ orthocentric system with $D$ on nine point circle. Let $U$ and $V$ be reflections of $D$ across $AH$ and $BC$. Let $P$ and $Q$ be the orthocenters of $\triangle ADH$ and $\triangle BDC$. Show that $PQUV$ form orthocentric system. (this suffices) (this is trivial by coordinate bash i think but i mean synthetic = yes) linpop seriously man

[asy][asy] size(11cm); defaultpen(fontsize(10pt)); pair A = dir(115), B = dir(205), C = dir(335), D = dir(40), M1 = dir(270), M2 = dir(90), I = incenter(A,B,C), Ia = 2*M1 - I, Ib = extension(A,M2,C,Ia), Ic = extension(A,M2,B,Ia), E = extension(A,2*foot(D,M1,M2)-D,B,C), E1 = 2M1-E, E2 = 2M2-E, A1 = extension(A,I,B,C), D1 = extension(E,E1,D,A1), D2 = extension(D,extension(A,Ib,B,C),E,M2), F = E + dir(E--D)*abs(E-D1)*abs(E-E1)/abs(E-D), D0 = 2*foot(D,M1,M2)-D; draw(circumcircle(A,B,C)); draw(arc(circumcenter(D,Ib,Ic), abs(D-circumcenter(D,Ib,Ic)), -140, 5), purple); draw(A--B--C--A); draw(Ia--Ib--Ic--Ia); draw(E1--E2^^M1--M2, red); draw(D--E, red+dashed); draw(B--E); draw(circumcircle(D,I,Ia), purple); draw(E1--E--E2, red); draw(A--Ia); draw(circumcircle(B,C,Ib)^^circumcircle(B,I,C), blue+dotted); draw(circumcircle(D1,E1,D2), purple+dashed); draw(A--E, dotted); dot("$A$", A, dir(115)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$I$", I, dir(180)); dot("$I_a$", Ia, dir(270)); dot("$I_b$", Ib, dir(0)); dot("$I_c$", Ic, dir(225)); dot("$E_1$", E1, dir(315)); dot("$E_2$", E2, dir(0)); dot("$D_1$", D1, dir(225)); dot("$D_2$", D2, dir(225)); dot("$M_1$", M1, dir(270)); dot("$M_2$", M2, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, dir(180)); dot("$F$", F, dir(90)); dot("$D'$", D0, dir(135)); [/asy][/asy] Let $\overline{AI}$ meet $(ABC)$ again at $M_1$, and $M_2$ be the midpoint of arc $BAC$. Let $(DII_a)$ meet $(ABC)$ again at $D_1$, and $(DI_bI_c)$ meet $(ABC)$ again at $D_2$. By radical axis on $(BICI_a)$, $(DID_1I_a)$, and $(ABC)$, we have $\overline{AI}$, $\overline{BC}$, $\overline{DD_1}$ concur. By radical axis on $(BCI_bI_c)$, $(DD_2I_bI_c)$, and $(ABC)$, we have $\overline{AM_2}$, $\overline{BC}$, $\overline{DD_2}$ concur. Thus $$-1 = (B,C;\overline{AI} \cap \overline{BC},\overline{AM_2} \cap \overline{BC})\overset D= (B,C;D_1,D_2).$$ Since $M_1,M_2$ are arc midpoints, we then have $\overline{D_1M_1}$, $\overline{D_2M_2}$, $\overline{BC}$ at some point $E$. We will show that $\angle BAD = \angle EAC$ and $D$, $E$, $F$ collinear. $\angle BAD = \angle EAC$: Let $D'$ be the reflection of $D$ over $\overline{M_1M_2}$. Let $f(P) = \pm \frac{PB}{PC}$ depending on if $P$ is above or below $\overline{BC}$. By Ratio lemma we have $$f(E) = f(D_1) = \frac{f(\overline{AI} \cap \overline{BC})}{f(D)} = f(A)f(D'),$$ so $A$, $D'$, $E$ are collinear which implies the conclusion. $D$, $E$, $F$ collinear: Let $E_i$ be the reflection of $E$ over $M_i$ for $i = 1,2$. By Shooting lemma we have $$M_1D_1 \cdot M_1E_1 = M_1D_1 \cdot M_1E = M_1B^2 = M_1I \cdot M_1I_a,$$ so $E_1 \in (DII_1)$ and similarly $E_2 \in (DI_bI_c)$. So since $\overline{M_1M_2} \parallel \overline{E_1E_2}$ we have $D_1E_1E_2D_2$ is cyclic by Reim's, so we are done by radical axis on $(DI_bI_c)$, $(DII_a)$, and $(D_1E_1E_2D_2)$. $\square$ MarkBcc168 wrote: Prove that the midpoint of $EF$ lies on $\odot(ABC)$. Note that $\angle E_1FE_2 = \angle E_1FD + \angle E_2FD = \angle DD_1M_1 + \angle DD_2M_2 = 90^\circ$, so if $X$ is the midpoint of $\overline{EF}$ then homothety centered at $E$ with scale factor $\frac 12$ implies $\angle M_1XM_2 = 90^\circ$.

We use directed angles and lengths. State the problem with reference to the orthocentric system $I,I_a,I_b,I_c$. Let $E'$ be the point on $\overline{BC}$ with $\measuredangle BAD=\measuredangle E'AC$. Notice that $(I,I_a)$ and $(D,E')$ are images under an inversion at $A$ with power $AB \cdot AC$ composed with a reflection over $\overline{AI}$, so we have $$AD \cdot AE'=AB \cdot AC=AI \cdot AI_a.$$ Let $X=\overline{AI} \cap \overline{DE'}$ and $P=\overline{DE'} \cap (ABC)$. Claim: $(DII_a)$ passes through $2P-E'$. Proof: It suffices to show $XD(2XP-XE')=XI \cdot XI_a$. By the incenter-excenter lemma, we have $AI+AI_a=2AM$, so $$\begin{align*} XI \cdot XI_a&=(XA+AI)(XA+AI_a) \\ &=XA^2+XA(AI+AI_a)+AI \cdot AI_a \\ &=XA^2+2XA \cdot AM+AD \cdot AE' \\ &=2XA(XA+AM)-XA^2+AD \cdot AE' \\ &=2XA \cdot XM-AX^2+AD \cdot AE'. \end{align*}$$ By Stewart on $ADE'$, we have $$\begin{align*} AX^2&=\frac{AD^2 \cdot E'X+AE'^2 \cdot DX}{DE'}+XD \cdot XE' \\ &=\frac{AD^2 \cdot \frac{AE' \cdot DE'}{AD+AE'}+AE'^2 \cdot \frac{AD \cdot DE'}{AD+AE'}}{DE'}+XD \cdot XE' \\ &=\frac{AD \cdot AE' \cdot DE'(AD+AE')}{DE'(AD+AE')}+XD \cdot XE' \\ &=AD \cdot AE'+XD \cdot XE'. \end{align*}$$ Plugging this into the expression for $XI \cdot XI_a$, we obtain $$XI \cdot XI_a=2XA \cdot XM-XD \cdot XE'=XD\left(\frac{2XA \cdot XM}{XD}-XE'\right)=XD(2XP-XE'),$$ as desired. $\square$ We can swap $I$ with $I_b$ and $I_a$ with $I_c$ and apply this claim to show $(DI_bI_c)$ passes through $2P-E'$ as well. Thus, $2P-E'=F$, so $E'=\overline{DF} \cap \overline{BC}=E$, as desired. $\square$ Remark: How do you guess $F=2P-E'$? You don't. Or, you can solve half of the problem synthetically in-contest and derive that it must be true but fail to solve the other half of the problem which is literally the same thing in the orthocentric system statement.

$(O)$ is the circumcircle of $ABC$. Redefine $E$ as the intersection of the isogonal line of $AD$ and $BC$. $H$ is the midpoint of arc $BAC$, $(DI_bI_c)$ intersect $(O)$ the second time at $G$, $I_bI_c$ cuts $BC$ at $J$ We have $I_bI_cBC$ is cyclic, therefore $JI_b.JI_c=JB.JC$, so by radical axis we have $J,D,G$ are collinear $AE$ cuts $(O)$ the second time $E'$, we have: $\displaystyle\frac{BE}{CE}=\frac{AB}{AC}.\frac{BE'}{CE'}=\frac{JB}{JC}.\frac{CD}{BD}=\frac{GB}{GC}$ Thus $G,E,H$ are collinear $K$ is the intersection of $HD$ and $BC$ We then have $HA.HJ=HB^2=HK.HD=HE.HG$, which means $KEGD,AKDJ,AEGJ$ are all cyclic So $A$ is the Miquel point of the complete quadrilateral $KEGD.HJ$ Denote $L$ the intersection of $KG$ and $DE$, $M$ is the center of $(KEGD)$ We have a well-known lemma: Given a complete quadrilateral $ABCD.EF$ with $ABCD$ cyclic, center of $(ABCD)$ is $O$, $AC$ intersect $BD$ at $G$, then $OG$ pass through the Miquel point of $ABCD.EF$ By Brokard's theorem, wwe have $L$ is the orthocenter of $HMJ$, by the Lemma $A,L,M$ are collinear, and by $ML\perp JH$ we have $A,L,M,R,I_a$ are collinear, $R$ is the midpoint of arc $BC$ not containing $A$. We have to prove that $D,E,F$ are collinear, which is equivalent to show that $D,L,F$ are collinear, which means $L$ lies on the radical axis of $(DII_a)$ and $(DI_bI_c)$ $(I_bI_cD)$ intersect $(BIC)$ at $P,Q$. We have $JI_b.JI_c=JB.JC$, thus $J,P,Q$ are collinear by radical axis Notice that if we prove that $P,L,Q$ are collinear, then $LP.LQ=\mathcal{P}_{L-(DI_bI_c)}=LI.LI_a=\mathcal{P}_{L-(DII_a)}$, which ends the problem. Now I show that this is true. Define $S$ to be the center of $(DI_bI_c)$. By Brokard's theorem, $JL\perp HM$, and we have $SR\perp PQ$, so we need to prove $SR\parallel HL$ $H$ is the midpoint of $I_bI_c$ so $HS\parallel MR$, and note that $S,O,M$ are collinear because they all lie on the perpendicular bisector of $DG$, and $O$ is the midpoint of $HR$, thus $HSRM$ is a parallelogram, this implies $HL\parallel SR$, the problem is proved

Here's a slick (scam) dot product solution: Redefine $E$ as the point on $BC$ satisfying $\angle BAD = \angle EAC$. Observe that by $\sqrt{bc}$ inversion, $\measuredangle IEI_a = -\measuredangle IDI_a$ and so if $I_aE$ intersects $(DII_a)$ again at $X$, then $IE = IX$. So, we get that the power of $E$ w.r.t. $(DII_a)$ equals $(EI)(EX) = 2 \overrightarrow{EI} \cdot \overrightarrow{EI_a}$ and similarly the power of $E$ w.r.t. $(DI_bI_c)$ is $2\overrightarrow{EI_b}\cdot \overrightarrow{EI_c}$ So, we want to now show that $\overrightarrow{EI}\cdot \overrightarrow{EI_a} = \overrightarrow{EI_b}\cdot \overrightarrow{EI_c}$. Writing this in terms of position vectors, this becomes $(e-i)\cdot (e-i_a)= (e-i_b)(e-i_c)$ or $e\cdot (i_b+i_c-i-i_a) = i_b\cdot i_c-i\cdot i_a$ which has to be the equation of a line in $e$. But as $E=B,C$ clearly satisfy $\overrightarrow{EI}\cdot \overrightarrow{EI_a} = \overrightarrow{EI_b}\cdot \overrightarrow{EI_c}=0$; $\overrightarrow{EI}\cdot \overrightarrow{EI_a} = \overrightarrow{EI_b}\cdot \overrightarrow{EI_c}$ in terms of $E$ represents the line $BC$ and we are done!

[asy][asy] import geometry; unitsize(3cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair L = dir(270); pair D = dir(240); triangle t = triangle(A,B,C); pair I = incenter(t); pair Ia = excenter(t.BC); pair Ib = excenter(t.CA); pair Ic = excenter(t.AB); pair E = intersectionpoint(line(B,C), reflect(line(A,I)) * line(A,D)); pair X = reflect(line(A,I)) * E; pair Y = reflect(line(Ib, Ic)) * E; draw(circumcircle(A,B,C), red); draw(circumcircle(D, I, Ia), green); draw(circumcircle(D, Ib, Ic), green); draw(circumcircle(B,I,C), blue); draw(circumcircle(B,C,Ib), blue); draw(A--B--C--cycle); draw(A--Ia); draw(E--X, dashed+orange); draw(E--Y, dashed+orange); draw(Ia--Ib--Ic--cycle); draw(A--D); draw(A--E); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$L$", L, dir(L)); dot("$I$", I, dir(I)); dot("$I_a$", Ia, dir(Ia)); dot("$I_b$", Ib, dir(Ib)); dot("$I_c$", Ic, dir(Ic)); dot("$E'$", E, dir(E)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); [/asy][/asy] Let $E'$ be the point on $BC$ such that $\angle BAD = \angle EAC$. I will show that $E'$ lies on the radical axis of $(DII_A)$ and $(DI_BI_C)$, which clearly proves $E = E'$. Claim: We have $$AD \cdot AE' = AB \cdot AC = AI \cdot AI_a = AI_b \cdot AI_c$$ Proof. The first equality follows from $\triangle ABD \sim \triangle AE'C$, which is simple angle chasing. The second equality follows from $\triangle AIB \sim \triangle ACI_a$, which is also simple angle chasing. The last equality follows from the fact that $I$ is the orthocenter of $\triangle I_aI_bI_c$ and $A$ is the foot from $I_a$ to $I_bI_c$, and then using the fact that the reflection of $I$ across $I_bI_c$ lies on $(I_aI_bI_c)$. $\square$ Claim: Let $X$ and $Y$ be the reflections of $E'$ across $II_a$ and $I_bI_c$, respectively. Then $DXII_a$ and $DYI_bI_c$ are cyclic. Proof. We have $$AI \cdot AI_a = AD \cdot AE' = AD \cdot AX$$ so by Power of a Point, $DXII_a$ is cyclic. Similarly, $$AI_b \cdot AI_c = AD \cdot AE' = AD \cdot AY$$ so by Power of a Point, $DYI_bI_c$ is cyclic as well. $\square$ Now, let $(DXII_a) = \Gamma_1$ and $(DYI_bI_c) = \Gamma_2$ for brevity. Let the circumcircles of $BICI_a$ and $BCI_bI_c$ be $\omega_1$ and $\omega_2$ respectively. Note that $\omega_1$ has diameter $II_a$ and $\omega_2$ has diameter $I_bI_c$. Claim: Let $f (\Omega, P)$ be the power of the point $P$ with respect to circle $\Omega$. Then $$f(\Gamma_1, E') - f(\omega_1, E') + f(\Gamma_1, X) - f(\omega_1, X) = 0$$ $$f(\Gamma_2, E') - f(\omega_2, E') + f(\Gamma_2, Y) - f(\omega_2, Y) = 0$$ and additionally $$f(\Gamma_1, X) = 0$$ $$f(\Gamma_2, Y) = 0$$ $$f(\omega_1, X) = f(\omega_1, E')$$ $$f(\omega_2, Y) = f(\omega_2, E')$$ $$f(\omega_1, E') = f(\omega_2, E')$$ Proof. The first two follow by Linearity of Power of a Point; in particular, $g(P) = f(\Omega_1, P) - f(\Omega_2, P)$ for fixed circles $\Omega_1$, $\Omega_2$ is a linear function with respect to $P$. Let the midpoint of $E'X$ be $M$, lies on $II_a$, the radical axis of $\Gamma_1$ and $\omega_1$. Then $$\frac{f(\Gamma_1, E') - f(\omega_1, E') + f(\Gamma_1, X) - f(\omega_1, X)}{2} = f(\Gamma_1, M) - f(\omega_1, M) = 0$$ which implies the first equation. Let the midpoint of $E'Y$ be $N$, which lies on $I_bI_c$, the radical axis of $\Gamma_2$ and $\omega_2$. Then $$\frac{f(\Gamma_2, E') - f(\omega_2, E') + f(\Gamma_2, Y) - f(\omega_2, Y)}{2} = f(\Gamma_2, N) - f(\omega_2, N) = 0$$ which implies the second equation. The following two equations follow from $X \in (DII_a)$ and $Y \in (DI_bI_c)$. The next two follow since $X$ is the reflection of $E$ across $II_a$, the diameter of $\omega_1$, and $Y$ is the reflection of $E$ across $I_bI_c$, the diameter of $\omega_2$. The last one follows since $BC$ is the radical axis of $\omega_1$ and $\omega_2$, and $E'$ lies on $BC$. $\square$ Then, it follows that $$\begin{align*} f(\Gamma_1, E') &= f(\omega_1, E') - f(\Gamma_1, X) + f(\omega_1, X) \\ &= f(\omega_1, E') - 0 + f(\omega_1, E') \\ &= f(\omega_2, E') - 0 + f(\omega_2, E') \\ &= f(\omega_2, E') - f(\Gamma_2, Y) + f(\omega_2, Y) \\ &= f(\Gamma_2, E') \end{align*}$$ which implies that $E'$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$, which is what we wanted to prove. $\blacksquare$