Slovenija 2008

Denote the intersection of lines $AB$ and $FG$ by $H$ and let $J$ be the point where the bisector of the angle $\angle BAC$ meets the segment $BC$. Since the line $FG$ is parallel to $AJ$, we have $\angle HGA=\angle FGD=\angle BAJ=\angle AHG$, so the triangle $GAH$ is isosceles and $|AH|=|AG|$. We know that $|BE|=|CD|$ and from what we just proved $|AH|=|AG|$. It remains to be seen that $|CG|=|AB|$. The figure reveals no similar triangles, so we have to find a different way to use the equality of lengths. Lengths play an important role in Ceva's and Menelaus's theorem. The former does not apply here, but we can try to use the latter.

Menelaus's theorem for the triangle $AEC$ and the line $GF$ states that $$\frac{|AH|}{|HE|}\cdot \frac{|EF|}{|FC|}\cdot \frac{|CG|}{|GA|}=1,$$ and since $|AH|=|AG|$ we get $\frac{|CG|}{|HE|}=\frac{|FC|}{|EF|}$. Now, let us try to express the ratio $\frac{|FC|}{|EF|}$ in another way. Menelaus's theorem for the triangle $AEC$ and the line $BF$ gives us $$\frac{|AB|}{|BE|}\cdot \frac{|EF|}{|FC|}\cdot \frac{|CD|}{|DA|}=1,$$ and since $|BE|=|CD|$ we have $\frac{|AB|}{|DA|}=\frac{|FC|}{|EF|}$. Hence, $\frac{|AB|}{|DA|}=\frac{|CG|}{|HE|}$. This is equivalent to $$\frac{|AE|+|EB|}{|AG|+|GD|}=\frac{|CD|+|DG|}{|AH|+|AE|},$$ which implies $|AE{|}^2+|AE|(|EB|+|AH|)=|GD{|}^2+|GD|(|AG|+|CD|)$. Since $|AG|+|CD|=|EB|+|AH|$, we have $$\begin{array}{rl}0& =|AE{|}^2-|GD{|}^2+|AE|(|EB|+|AH|)-|GD|(|EB|+|AH|)\\ & =(|AE|-|GD|)(|AE|+|GD|+|EB|+|AH|),\end{array}$$ so $|AE|=|GD|$, and, finally, $|AB|=|AE|+|EB|=|GD|+|DC|=|GC|$, which was to be shown.