Ok, this is unnecessarily long. In the actual contest, I found a solution using moving points, had some time left, scratched a figure on a napkin, and found this solution : Consider $f$ the reflection over line $BC$. What we want to show is: $f(CF)=CG$. Let $f(X)=X’$. We have $F’ \in (BA’)$ and $EF’$ is parallel to $BD’$. But $BD=DC=BD’=CD’$, thus $DC$ is parallel to $BD’$, which is in turn parallel to $EF’$. We want to show that $F’$ lies on $CG$. The problem becomes: Prove that $CG,BA’$ and the line through $E$ parallel to $AC$ are concurrent. Let $P=(CG) \cap (BA’)$. Let $X=(PE) \cap (AC)$. Now, by simple angle chase $A’C$ is parallel to $BD$. By Desargues’ theorem on $\triangle A’CX$ and $\triangle BGE$, we have that $A’C \cap BG = P_{\infty}, A, (A’X) \cap (BE)=M$, are collinear, or $(AM)$ is parallel to $BD$. Let $N$ be the intersection of the line through $E$ parallel to $AC$ and line $BC$, by angle chase $A’N=A’C=AC$, thus $ANA’C$ is a parallelogram, and $AN$ is parallel to $A’C$ which is in turn parallel to $BD$, thus: $M=N$, and $A’X \equiv A’M$ is parallel to $AC$, which implies $X= P_{\infty}(AC)$, as desired.

I also found a projective solution. Given that this is a PAMO #1, I hope that a non-projective solution also exists. Let $H = AB\cap CE$, $Q = EF\cap AC$, $R = CG\cap EF$, and $P$ be the point on $EF$ with $PC\perp CE$. Because $QE = QC$, point $Q$ is the midpoint of $\overline{PE}$. Then \[ -1 = (PEQ\infty) \stackrel{D}=(XECB), \]where $X = PD\cap BC$. It follows by uniqueness of the cross-ratio that $P$, $D$, $H$, and $X$ are collinear. Finally, \[ -1 = (XECB) \stackrel{H}=(PERF), \]and because $\angle PEC = 90^\circ$ we deduce $\angle ECR = \angle ECF$.

A very easy problem, set FE intersection AC at point L, then ∠GDC∽⊿CLF

Let $H$ be the refection of $G$ across the perpendicular bisector of $BC$ then evidently $H\in AC$ and $GH\parallel BC$. Now $$\frac{AB}{AF}=\frac{AG}{AE}=\frac{AH}{AC}$$Hence $BH\parallel FC$. Finally $\angle GCB= \angle HBC= \angle BCF$.

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This was short Let $H$ be the intersection of that parallel line with $AC$ Then the given condition is equivalent to prove $\angle HCT = \angle HFC$ which is equivalent to prove $HC$ is tangent to $(TCF)$ where $T$ is $GC \cap HF$, i.e. $HC^2=HF*HT$ $\leftrightarrow \frac{BD*AH}{AD} * \frac{GD*CH}{CD}=HC^2$ which simplifies to:(after some length chasing with parallel lines) $\leftrightarrow EH=HC$ which is true

By the trigonometric ceva theorem we can solve this problem

Okay, here's another one. Let $Y$ be a point such that $ABCY$ is isosceles trapezoid. It's easy to show that $Y$ lies on $BD$. We will show that $\bigtriangleup FAC\sim\bigtriangleup CYG$. Note firstly that by the trapezoid we have that $\angle CAF=\angle CYG$. Since $AY\parallel BE$ we have that $\frac{GE}{AG}=\frac{GB}{YG}\Rightarrow \frac{GE}{AG}+1=\frac{GB}{YG}+1=\frac{AE}{AG}=\frac{YB}{YG}$. Since $BG\parallel FE$ we have that $\frac{YB}{YG}=\frac{AE}{AG}=\frac{AF}{AB}$. Also, since $YB=AC$ and $YC=AB$, we have that $\frac{AF}{YC}=\frac{AC}{YG}\Rightarrow \frac{AF}{AC}=\frac{YC}{YG}$. So by SAS we showed that $\bigtriangleup FAC\sim\bigtriangleup CYG$. Now, let $X=FG\cap BE$. Since $FE\parallel BG\Rightarrow\frac{FX}{XG}=\frac{FE}{BG}=\frac{AF}{AB}=\frac{AF}{CY}=\frac{CF}{CG}$, and we are done by Angle Bisector Theorem.

Let the line through $F$ parallel to $AC$ meet line $BC$ at $K$. Then $\triangle KFB\sim\triangle CAB$ and $\dfrac{BC}{BK}=\dfrac{AB}{FB}$. Also, we have $\angle FEB=\angle GBC=\angle ACB=\angle FKC$, so $\triangle KFE$ is isosceles with $KF=FE$. Since $BD$ and $EF$ are parallel, we see that \[\dfrac{KC}{BC}=1+\dfrac{KB}{BC}=1+\dfrac{FB}{AB}=\dfrac{AF}{AB}=\dfrac{EF}{BG}=\dfrac{KF}{BG}.\]Thus, we get $\dfrac{KC}{KF}=\dfrac{BC}{BG}$, so $\triangle GBC\sim\triangle FKC$. Hence, we get $\angle BCG=\angle KCF=\angle BCF$.

DDIT at $C$ on $BGEF$.

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kerryberry wrote: In a triangle $ABC$ with $AB<AC$, $D$ is a point on segment $AC$ such that $BD = CD$. A line parallel to $BD$ meets segment $BC$ at $E$ and line $AB$ at $F$. Point $G$ is the intersection of $AE$ and $BD$. Show that $\angle BCG = \angle BCF$. (Côte d’Ivoire) let $H$ be the intersection of lines $FE$ and $CD$, and we know that $BD\parallel HF$, so we get that $\angle HEC= \angle HCE$ and $\angle ABD=\angle AFH$, since $\angle HCE $ is an external angle for $\triangle CEF$ we can show that $\angle EFC= \angle 3+\angle 4-\angle 6\implies\angle 7=\angle 3+\angle 4-\angle 6$ now by trigonometric ceva we get \begin{align*} \frac{\sin(\angle 1)}{\sin(\angle 5)}\cdot\frac{\sin(\angle 3+\angle 4)}{\sin(\angle4)}\cdot \frac{\sin(\angle3)}{\sin(\angle2)}=1=\frac{\sin(\angle1)}{\sin(\angle5)}\cdot \frac{\sin(\angle3+\angle4-\angle6)}{\sin(\angle6)}\cdot\frac{\sin(\angle4+\angle3)}{\sin(\angle2)} \end{align*} \begin{align*} &\frac{\sin(\angle3)}{\sin(\angle4)}=\frac{\sin(\angle3+\angle4-\angle6)}{\sin(\angle6)}\\ &\sin(\angle6)\cdot\sin(\angle3)=\sin(\angle4)\cdot\sin(\angle3+\angle4-\angle6)\\ &\frac{\cos(\angle6-\angle3)-\cos(\angle6+\angle3)}{2}=\frac{\cos(\angle6-\angle3)-\cos(\angle3+2\times\angle4-\angle6)}{2}\\ &\text{multiplying both sides by $2$ we get }\\ &\cos(\angle6-\angle3)-\cos(\angle6+\angle3)=\cos(\angle6-\angle3)-\cos(\angle3+2\times\angle4-\angle6)\\ &\cos(\angle3+2\times\angle4-\angle6)=\cos(\angle6+\angle3)\implies \angle6+\angle3=\angle3+2\times\angle4-\angle6\implies \angle4=\angle6 \end{align*}hence we are done

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another solution. let $X$ be a point on $BC$ such that $\measuredangle CYF=\measuredangle DBC$, let $Y$ be a point on line $XF$ such that $\measuredangle XCY=\measuredangle CXY$ $$\boxed{\textbf{Claim:}\triangle FYC\sim\triangle AGD}$$by doing some work we find that the claim is true

Let $R=EF\cap AC$, $S=FG\cap BC$, and $T=BD\cap CF$. Note that $\triangle CGD\sim \triangle FCR$. Hence, $$\frac{CG}{CF}=\frac{GD}{CR}=\frac{GD}{ER}=\frac{BG}{FE}=\frac{GS}{SF},$$so we are done by the angle bisector theorem.

This can also be done by Trig and using Thales Theorem couple of times I am too lazy to write up the solution in but here is it my solution I wrote in on paper Also I didn't mention there but in the end we are done because from a well known lemma we know that: $$\frac{\sin x}{\sin z - \sin x}=\frac{\sin y}{\sin z - \sin y} \implies x=y$$ So $\frac{\sin \angle BCG}{\sin C - \angle BCG}=\frac{\sin \angle FCE}{\sin C - \angle FCE} \implies \angle BCG=\angle FCE \equiv FCB \equiv BCF \implies \angle BCG=\angle BCF \blacksquare$

Attachments:
PAMO 2023 P1.pdf (325kb)