BO=2DO(BO is twice as much as DO),
The midline AF on the side of BC must pass through point O.
Prove: let OB midpoint m and OC midpoint n,
Even MN, ND, DE, EM,
By 200 BC, Germany was half that of BC,
MN‖BC, MN is half of BC,
∴DE‖MN, and DE=MN,
∴ quadrilateral EMND is a parallelogram.
The two diagonal lines MD and EN are equally divided,
∴OM=OD,∴BO=2DO,
Similarly: ON=OE,∴CO=2EO.
O is the center of gravity of triangle ABC,
∴AF is the midline of BC, which must pass through O.
(The intersection of the three midlines is the center of gravity of the triangle).