(a) First, note that a composition of a reflection about the $A$-angle bisector and a homothety sends $\triangle AEF$ to $\triangle ABC$. It is well-known that $I$ is the midpoint of $AH$ because $AH$ is the diameter of cyclic quadrilateral $AFHE$. Letting $A'$ be the point diametrically opposite $A$ in the circumcircle of $\triangle ABC$, we see that $H$ is sent to $A'$. Since $J$ is sent to $K$, this means that $G$ is sent to $KA'\cap (ABC)$, which is well-known to be $(ABC)\cap (AFE)$. Suppose this intersection point is $G'$. Then $GA$ and $G'A$ are reflections about the $A$-angle bisector, which means that $\widehat{G'F}=\widehat{GE}$, which means that $GG'\parallel FE$. Since $KF=KE$ ($K$ is the circumcenter of cyclic quadrilateral $BFEC$), this means that by symmetry, $HL\parallel EF$. Supposing that $L$ is sent to $L'$, then $L'A'\parallel BC$, which means that $L'$ is the reflection of the orthocenter of $\triangle ABC$. The conclusion follows. (b) The following rephrasing using different letters might be helpful: VMO 2021 P3(b), rephrased wrote: Let $\triangle ABC$ be a triangle with circumcenter $O$. Suppose $D$ is the foot of the perpendicular from $A$ to $BC$. Let $OD$ intersect $(BOC)$ at $T$ and $AO$ intersect $(BOC)$ at $X$. Denote the intersection points of $TX$ with $AB$ and $AC$ by $P$ and $Q$. Show that $BQ\cap CP$ lies on the $A$-symmedian. This could be fairly easily finished with barycentric coordinates (the only hard part is computing $T$ and $X$). Anyone has synthetic observations for this part?

@above: Here is my solution for what you are left. By Ceva-Menelaus configuration, it suffices to show that $AA, BC, TX$ are concurrent. Let $\gamma$ denote the $A$-Apollonius circle, which is centered at $AA\cap BC$ and is orthogonal to $\odot(ABC)$. Let $K = AO\cap BC$. Notice that $D,K$ are inverses w.r.t. $\gamma$. Thus, by inverting this relation around $\odot(ABC)$, we find that $T,X$ are inverses w.r.t. $\gamma$. This gives the desired result.

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/* end of picture */[/asy][/asy] Part a) We easily have the tagents at $F,E$ WRT $(I)$ intersect at thi midpoint of $BC$ so $GFLE$ is harmonic. $\implies H(LG,EF)=-1\implies H(LJ,EF)=-1$ but $J$ is the midpoint of $EF$ so $HL\parallel FE$ which claims $AL\bot EF$ since $AL\bot LH$. $\quad\blacksquare$ Part b) (not mine) By power point $M$ we have $MI.MN=ME.MF=MA.ML$ so easy to see that $\angle AND=90^o$ so we claim that $AA$ WRT $(I),ND,EF$ are concurrent at $S$ by radical axis of $(AD), \text{Euler circle}\ , (I)$ Consider $\bigtriangleup APQ$ has $\overline{FE}\cap\overline{PQ}=S$ and since $AK$ is the symmedian of $\bigtriangleup AFE$ we have $$-1=A(S,\overline{AK}\cap\overline{FE},E,F)=A(S,\overline{AK}\cap\overline{PS},Q,P)\implies AK,PE,QF\quad\text{are concurrent}.\quad\blacksquare$$

a. It's well known that KF and KE are both tangent to $(I)$. Hence, $GL$ is symmedian of $\triangle FGE$. Therefore, \[\angle FAL=\angle FGL=\angle MGE=\angle HGE=\angle HAE=90^{\circ}-\angle AHE=90^{\circ}-\angle AFE\]which implies $AL\perp EF$. b. Define $X=PQ\cap EF$, $Y=PE\cap QF$, $Z=AY\cap EF$, $T=AH\cap FE$. Define also $X'$ as the intersection of $EF$ and line parallel to $BC$ through $A$. Notice that \[(X,M;F,E)\stackrel{N}{=}(D,I;F,E)\stackrel{I}{=}(T,P_{\infty};F,E)=A(T,P_{\infty};F,E)\]Moreover, from reflection across the internal bisector of $\angle BAC$, we have \[A(T,P_{\infty};F,E)=A(M,X';E,F)=(M,X';E,F)=(X',M;F,E)\]$(X,M;F,E)=(X',M;F,E)$, so $X=X'$. Hence, $AX\parallel BC$. Now, notice that $-1=(X,Z;P,Q)\stackrel{A}{=}(P_{\infty},AZ\cap BC;B,C)$. Therefore, we must have $K=AZ\cap BC$ which finish the problem.

ACGNmath wrote: (a) First, note that a composition of a reflection about the $A$-angle bisector and a homothety sends $\triangle AEF$ to $\triangle ABC$. It is well-known that $I$ is the midpoint of $AH$ because $AH$ is the diameter of cyclic quadrilateral $AFHE$. Letting $A'$ be the point diametrically opposite $A$ in the circumcircle of $\triangle ABC$, we see that $H$ is sent to $A'$. Since $J$ is sent to $K$, this means that $G$ is sent to $KA'\cap (ABC)$, which is well-known to be $(ABC)\cap (AFE)$. Suppose this intersection point is $G'$. Then $GA$ and $G'A$ are reflections about the $A$-angle bisector, which means that $\widehat{G'F}=\widehat{GE}$, which means that $GG'\parallel FE$. Since $KF=KE$ ($K$ is the circumcenter of cyclic quadrilateral $BFEC$), this means that by symmetry, $HL\parallel EF$. Supposing that $L$ is sent to $L'$, then $L'A'\parallel BC$, which means that $L'$ is the reflection of the orthocenter of $\triangle ABC$. The conclusion follows. (b) The following rephrasing using different letters might be helpful: VMO 2021 P3(b), rephrased wrote: Let $\triangle ABC$ be a triangle with circumcenter $O$. Suppose $D$ is the foot of the perpendicular from $A$ to $BC$. Let $OD$ intersect $(BOC)$ at $T$ and $AO$ intersect $(BOC)$ at $X$. Denote the intersection points of $TX$ with $AB$ and $AC$ by $P$ and $Q$. Show that $BQ\cap CP$ lies on the $A$-symmedian. This could be fairly easily finished with barycentric coordinates (the only hard part is computing $T$ and $X$). Anyone has synthetic observations for this part? Why can b) be rephrased this way??

Part (a) : $BCEF$ is cyclic with center $K \implies KJ \perp EF$ Also, $IJ \perp EF \implies I,J,K$ are collinear. $JI.JK=JF.JE=JG.JH \implies IHGK$ cyclic. $\angle HIK = \angle HGK = \angle HGL = \angle HAL \implies IK \parallel AL$ So, $AL \perp EF$ Part (b) same as DNCT1.

a) It's easy to see that $GK$ is $G$ - symmedian of $\triangle GEF,$ so $GH, GL$ are isogonal conjugate in $\angle{EGF}$. Then $\angle{LAE} = \angle{LGE} = \angle{HGF} = \angle{HAF}$ or $AL \perp EF$ b) Let $T$ be intersection of tangent at $A$ of $(AEF)$ with $EF,$ $DT$ intersects $(DEF)$ again at $N',$ $AH$ intersects $EF$ at $U$. We have $\dfrac{N'E}{N'F} = \dfrac{TE}{TF} \cdot \dfrac{DF}{DE} = \dfrac{AE^2}{AF^2} \cdot \dfrac{UF}{UE} = \dfrac{AE^2}{AF^2} \cdot \dfrac{UF}{UE} = \dfrac{ME}{MF}$. So $N'M$ is bisector of $\angle{EN'F}$ or $N', M, I$ are collinear. Hence $N' \equiv N$. From this, if we let $V \equiv PE \cap QF$ then $A(PQ, VT) = - 1$. But $A(PQ, KT) = A(FE, KT) = - 1$ so $A, V, K$ are collinear or $AK, PE, QF$ concur at $V$