Školjka

%V0 Let $ABC$ be a triangle, and let the angle bisectors of its angles $CAB$ and $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $F$, respectively. The lines $AD$ and $BF$ meet the line through the point $C$ parallel to $AB$ at the points $E$ and $G$ respectively, and we have $FG = DE$. Prove that $CA = CB$. Original formulation: Let $ABC$ be a triangle and $L$ the line through $C$ parallel to the side $AB.$ Let the internal bisector of the angle at $A$ meet the side $BC$ at $D$ and the line $L$ at $E$ and let the internal bisector of the angle at $B$ meet the side $AC$ at $F$ and the line $L$ at $G.$ If $GF = DE,$ prove that $AC = BC.$