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
Also it's clear that AH || OM since both are in the direction of B+C.
H.P.

Geometric proof:
AH || OM since both are perpendicular to BC. Then extend BO s.t. BO = OD.
=> BOD is a diameter of the circumcircle and D lies on the circumcircle.
By Thales', BAD = BCD = 90 degree.
=> AD perpendicular to AB and CD perpendicular to BC.
CH also perpendicular to AB => AH || CD
OM || DC
So AHCD is a ||gm.
AH = DC
In triangle BDC using midpoint theorem OM || DC and OM = DC/2.
=> OM  = AH/2 H.P.



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