69th Belarusian Mathematical Olympiad

Denote the points of intersection of the line $EG$ with the lines $BD$ and $CF$ by $P_1$ and $P_2$, respectively. We will prove that $P_1\equiv P_2$. Denote the points of intersection of the lines $AB$ and $AC$ with the line $DF$ by $C_1$ and $B_1$, respectively. The segment $A{A}_1$ is the altitude and the bisector in the triangle $A{B}_1{C}_1$, hence $△A{B}_1{C}_1$ is isosceles with the base $B_1{C}_1$ and $A_1{B}_1=A_1{C}_1$. Without loss of generality, suppose that $\angle A{A}_{1}B>\angle A{A}_{1}C$. If $A_1{B}_1\le A{A}_1$, the lines $BD$ and $CF$ intersect the ray $AE$, and if $A_1{B}_1>A{A}_1$, these lines intersect the ray $AG$. Therefore, it is enough to prove the equality $A{P}_1=A{P}_2$.

Suppose that $A_1{B}_1\le A{A}_1$ (the case $A_1{B}_1>A{A}_1$ can be considered similarly). Then $B_{1}F=A_{1}F-A_1{B}_1=A_{1}D-A_1{C}_1=D{C}_1$. From $DF\parallel EG$ follow the similarities $△B{C}_{1}D\sim △BA{P}_1$ and $△C{B}_{1}F\sim △CA{P}_2$, whence $$A{P}_1=\frac{AB\cdot D{C}_1}{B{C}_1}\text{and}A{P}_2=\frac{AC\cdot B_{1}F}{C{B}_1}.$$

Therefore, it is enough to prove that $AB/B{C}_1=AC/C{B}_1$. Let $C_2$ be the reflection of $C$ through the line $A{A}_1$. Since $△A{B}_1{C}_1$ is isosceles, the point $C_2$ lies on the segment $A{C}_1$. From the equalities $\angle C_2{A}_1{C}_1=\angle C{A}_1{B}_1=\angle C_1{A}_{1}B$ it follows that the line $A_1{C}_1$ is the bisector in the triangle $A_{1}B{C}_2$. Finally, $$\frac{C{B}_1}{B{C}_1}=\frac{C_1{C}_2}{B{C}_1}=\frac{A_1{C}_2}{A_{1}B}=\frac{A_{1}C}{A_{1}B}=\frac{AC}{AB}.$$