We are given $n$ circles which have the same center. Two lines $D_1,D_2$ are concurent in $P$, a point inside all circles. The rays determined by $P$ on the line $D_i$ meet the circles in points $A_1,A_2,...,A_n$ and $A'_1, A'_2,..., A'_n$ respectively and the rays on $D_2$ meet the circles at points $B_1,B_2, ... ,B_n$ and $B'_2, B'_2 ..., B'_n$ (points with the same indices lie on the same circle). Prove that if the arcs $A_1B_1$ and $A_2B_2$ are equal then the arcs $A_iB_i$ and $A'_iB'_i$ are equal, for all $i = 1,2,... n$.