Prove AH = 2OM pending

Prove that in a triangle ABC, AH = 2OM where H is the Orthocenter, O is the Circumcenter and M = midpoint of BC.


Using Sylvester's Triangle Theorem
We know that OH = OA + OB + OC so assume Circumcenter O as the origin of the vector space.
H = A + B + C
AH = OH - OA = B + C
OM = (OB + OC)/2 = (B + C)/2 proof here.
=> AH = 2OM
H.P.

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