AH = 2RcosA and OM1 = RcosA pending

Why is AH = 2RcosA in a triangle ABC and OM1 = R.cosA.
R = circumradius
H = orthocenter
O = circumcenter
M1 = midpoint of BC.

Proof for OM1 = R.cosA:
Since M1 is the midpoint of BC, OM1 will be perpendicular to BC.
BC is a chord of the circumcircle and hence it will subtend angle 2A at the center which will be bisected by OM1.
In triangle OM1C, OC = R and OC.cos(ROM1) = OC.cos(2A/2) = R.cosA.
Hence proved.


Proof for 
AH = 2R.cosA



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