Problem

Source: Gillis Olympiad 5777 (Israel National '16-'17) #5

Tags: national olympiad, geometry, pentagon, circumcircle



A regular pentagon $ABCDE$ is given. The point $X$ is on his circumcircle, on the arc $\overarc{AE}$. Prove that $|AX|+|CX|+|EX|=|BX|+|DX|$. Remark: Here's a more general version of the problem: Prove that for any point $X$ in the plane, $|AX|+|CX|+|EX|\ge|BX|+|DX|$, with equality only on the arc $\overarc{AE}$.