Archimedes' broken string theorem ":AB and BC are two strings of ⊙O (that is, ABC is a broken string of a circle), BC >；; AB and M are the midpoint of arc ABC, so the vertical foot G from M to BC is the midpoint of broken string ABC, that is, CG=AB+BG.

Extend the intersection o of MO to q

Then q is the midpoint of the broken string APB.

A vertical line intersecting with Q direction BP intersects with point E, and a vertical line intersecting with M direction BP intersects with point F..

So AP+PE=BE

So AP+BP=2BE.

Yi Zhi BM = bq, < MBQ=90.

So the triangle MBF is equal to the triangle BQE.

So BE=PF

Because I am the heart of the triangle headquarters base,

So PM is divided equally.

So PF+MF= root 2PM.