Given two circles $C_1,C_2$ with common center, the radius of $C_2$ is twice the radius of $C_1$. Quadrilateral $A_1A_2A_3A_4$ is inscribed in $C_1$. The extension of $A_4A_1$ meets $C_2$ at $B_1$; the extension of $A_1A_2$ meets $C_2$ at $B_2$; the extension of $A_2A_3$ meets $C_2$ at $B_3$; the extension of $A_3A_4$ meets $C_2$ at $B_4$. Prove that $P(B_1B_2B_3B_4)\ge 2P(A_1A_2A_3A_4)$, and in what case the equality holds? ($P(X)$ denotes the perimeter of quadrilateral $X$)
Problem
Source: China Mathematical Olympiad 1988 problem2
Tags: geometry, perimeter, geometry unsolved