TR-MO 1997 Problem 2

Let $F$ be a point inside a convex pentagon $ABCDE$ , and let $a_{1}$ , $a_{2}$ , $a_{3}$ , $a_{4}$ , $a_{5}$ denote the distances from $F$ to the lines $AB$ , $BC$ , $CD$ , $DE$ , $EA$ , respectively. The points $F_{1}$ , $F_{2}$ , $F_{3}$ , $F_{4}$ , $F_{5}$ are chosen on the inner bisectors of the angles $A$ , $B$ , $C$ , $D$ , $E$ of the pentagon respectively, so that $AF_{1} = AF$ , $BF_{2} = BF$ , $CF_{3} = CF$ , $DF_{4} = DF$ and $EF_{5} = EF$ . If the distances from $F_{1}$ , $F_{2}$ , $F_{3}$ , $F_{4}$ , $F_{5}$ to the lines $EA$ , $AB$ , $BC$ , $CD$ , $DE$ are $b_{1}$ , $b_{2}$ , $b_{3}$ , $b_{4}$ , $b_{5}$ , respectively. Prove that $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} \leq b_{1} + b_{2} + b_{3} + b_{4} + b_{5}$