Proof-the Concurrency of medians of a triangle

Concurrency of Medians of a Triangle Theorem--The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.
 Proof:

Since CD bisects AB at point D and BE bisects AC at point F.
Point D and F are the midpoints
Connect D and F, giving us the midsegment of △ABC.
Using the Midsegment theorem
DF∥BC and DF=BC/2
DF/BC=1/2
Since DF∥BC and they're cut by transversal BF.
∠DFB≅∠FBC by the alternate interior angles theorem.
∠BOC≅∠DOF by the vertical angles theorem.
△DOF∼△COB By the AA Similarity Postulate.
DF/BC=FO/BO=DO/CO
Since BF/BC=1/2
FO/BO=1/2
Since BO+FO=BF
BF-BO=FO
(BF-BO)/BO=1/2
BF/BO-BO/BO=1/2
BF/BO-1=1/2
BF/BO=3/2
BO/BF=2/3 (Reciprocal Property)
BO=2BF/3 (Prove)
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