Define $AG\cap BC=H, FE\cap AG=I, GE\cap AF=J$. Claim 1: $E$ is the orthocentr of $AFG$. Proof. Note that $$\angle BEC=\frac{1}{2}(360-\angle BDC)=\frac{1}{2}(180+2\angle A)=90+\angle A\implies \angle IEG=90-\angle A\implies \angle EIG=90.$$Similarly, $\angle FJE=90$. This means $E$ is the orthocenter of $AFG$. Claim 2: $AEHC$ is cyclic. Proof. Note that $$\angle GCH=\frac{1}{2}\angle EDB=90-\angle DEB=90-\angle AEI=\angle GAE.$$ Claim 3: The tangent at $A$ to $(AFG)$ is parallel to $BC$. Take a point $T$ on the parallel line to $BC$ through $A$ on the "right" of $A$. Then, $$\angle TAG=\angle AHB=180-\angle AHC=180-\angle AEG=\angle AFG.$$ This concludes the proof.

config meme-ing, well kind of. Let $Q= \overline{AD} \cap (BDC)$, note that $\measuredangle BQA = \measuredangle BFA$ and $\measuredangle CQA = \measuredangle CGA$ hence $FAQB$ and $CAQG$ are cyclic, POP at $E$ gives $FBGC$ cyclic and using $D$ is the circumcenter of $\triangle BEC$, angle chasing gives $GE \perp AF$ and $FE \perp AG$, thus $E$ is the orthocenter of $\triangle AFG$ which means tangent at $A$ to $(AFG)$ is parallel to $XY$ where $X$,$Y$ are the foot of perpendiculars from $G$ and $F$ onto $\overline{AF}$ and $\overline{AE}$ respectively. $\measuredangle YXG = \measuredangle YFG = \measuredangle BFG = \measuredangle BCG$ hence $XY \parallel BC$ and we are done.

MrOreoJuice wrote: on another note what is SMO? Singapore Math Olympiad (Open) Round 2, a 5 problem 4 hour paper of which this is q1 (q2 is the rectangle, q3 FE, q4 cards, q5 poly NT)

[asy][asy] size(12cm); pair A,B,C,D,E,F,G,H,K,L,P; A=(-5.26599,3.59133);B=(-4.18,-1.31);C=(2.3,-1.29);D=(-0.93,-4.504); E=(-3.082,-0.487);F=(-6.226,-2.843);G=(-2.101,-0.633); K=(-9.155,3.579);L=(7.245,3.63);P=(-3.632,0.54); draw(A--B--C--cycle);draw(K--D--L);draw(A--D);draw(F--E--C);draw(K--L,dashed); draw(circumcircle(A,B,C)); draw(circumcircle(A,B,F)); draw(circumcircle(A,C,G)); draw(circumcircle(E,B,C)); draw(circumcircle(A,F,G),dotted); label("$A$",A,dir(125)); label("$B$",B,dir(180)); label("$C$",C,dir(330)); label("$D$",D,S); label("$E$",E,dir(165)); label("$F$",F,dir(300)); label("$G$",G,dir(30)); label("$K$",K,dir(155)); label("$L$",L,dir(0)); label("$P$",P,dir(20)); dot(D);dot(E);dot(F);dot(G); [/asy][/asy] We notice immediately from the angle conditions that $\angle AFE=\angle BAC \implies (ABF)$ is tangent to $AC$ and similarly $(ACG)$ is tangent to $AB$. We then let $P=(ABF)\cap (ACG)$. Claim 1: $P$ lies on $AD$. Proof. We perform an inversion centered at $A$ with radius $\sqrt{AB\cdot AC}$, followed by a reflection about the bisector of $\angle BAC$. This swaps $B$ and $C$. Since $AD$ is the $A$-symmedian of $ABC$, $A'D'$ is the $A'$-median of $A'B'C'$. $P$ is sent to the point $P'$ such that $A'B'P'C'$ is a parallelogram, which is the reflection of $A'$ about the midpoint of $B'C'$. Therefore, $A',D',P'$ are collinear $\implies$ $A,D,P$ are collinear. So $D$ lies on the radical axis of $(ABF)$ and $(ACG)$. Then let $DB\cap (ABF)=K$, $DC\cap (ACG)=L$. $DB\cdot DK=DC \cdot DL\implies DK=DL$ Thus $KL//BC.$ Claim 2: $K,A,L$ are collinear. Proof: We first show that $B,D,C,P$ are concyclic, which follows from the following angle chase: $$\angle BPC=\angle BPE\,+\,\angle CPE=\angle AFE\,+\,\angle AGE=2\angle BAC=180^{\circ}-\,\angle BDC.$$Then the claim follows from the converse of Miquel's theorem applied to circles $(PBKA),(PALC),(PCDB)$. Claim 3: $KL$ is tangent to $(AFG)$. Proof: First note that $B,F,C,G$ are concyclic since $EB\cdot EF=EG\cdot EC.$ Then we finish with a straightforward angle chase: $$\angle AFG=\angle AFE\,+\,\angle EFG=\angle BCG\,+\,\angle BAC=\angle BCG\,+\,\angle BCD=\angle GCD=\angle GAL$$and we are done. Fun fact: The above proof generalises to show that the result holds for any point E chosen on the symmedian.

First, note by angle chasing that $\odot(AFB)$, $\odot(AGC)$, and $\odot(BDC)$ meet at point $P$. This motivates inversion at $E$ switching $\{A,P\}$, $\{B,F\}$, $\{C,G\}$. Let $D'$ be the image of $D$, we see that $D'\in\odot(AFG)$. Moreover, the condition $DB=DC=DE$ translates to that $ED'$ is the perpendicular bisector of $FG$, implying that $E$ is the orthocenter of $\triangle AFG$. Now, notice from inversion that $B,C,F,G$ are concyclic. Thus, the tangent to $\odot(EFG)$ at $E$ is parallel to $BC$. However, $\odot(EFG)$ and $\odot(AFG)$ are just translates with vector $\overrightarrow{EA}$, done.

Let $(BDC) \cap AE=T$ Easy to see $ATBF$ and $ATGC$ are cyclic. From $\angle AGE= \angle GEP + 90$ we see that $FE \perp AG$ and similarly $E$ is orthocenter of $AFG$. From $PoP$ from $E$ we can see $FBGC$ is cyclic too. Let perpendiculars from $F,G$ are $M,N$. Easy to see tangent $(AFG)$ from $A$ is parallel to $MN$. And rest of the solution is $\angle GNM = \angle GFM= \angle GCB$ so $MN \parallel BC \parallel AA$. question done

Let $(AFB) \cap DB=P$ and $(AGC) \cap DC=Q.$ Angle-chasing implies that $AP || BC, AQ || BC \implies P,A,Q$ are collinear. Then $\angle QPB = \angle CBD = \angle BCD \implies PQCB$ cyclic $ \implies D$ lies on radical axis of $(APFB)$ and $(AQCG).$ If $X=(APFB) \cap (AQCG)$, then $X \in AD.$ So, $EB \cdot EF = EX \cdot EA = EG \cdot EC \implies FBGC$ is cyclic. Simple angle-chasing yields $\angle AFG=\angle GCD= \angle QAG \implies PQ$ is tangent to $(AFG) \ \ \ \ \ \ \blacksquare$