Math IOQM

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.

Prove that in a triangle ABC with circumcenter O, OM = (OB + OC)/2 where OM,OB,OC are vectors. M is midpoint of BC. Proof: OM  = OB + BM = OB + (BC)/2 BC = OC - OB => OM = OB + OC/2 - OB/2 = (OB+OC)/2 H.P.

Prove that H = A + B + C in triangle ABC where Circumcenter O is the origin for the vector space and H,A,B,C are position vectors of the Orthocenter and vertices A,B,C respectively. Proof: For position vectors a,b,c: |a| = |b| = |c| = R = circumradius And a.a = b.b = c.c = R^2 Let a vector h = a + b + c. We will show that line segment from any vertex to H is perpendicular to the opposite side. AH = OH - OA = h - a = a + b + c - a = b + c BC = OC - OB = c - b Compute dot product of AH and BC AH.BC = (b + c) ( c - b) = b.c + c.c - b.b - c.b = c.c - b.b + b.c - c.b Since b.c = c.b AH.BC = c.c - b.b = R^2 - R^2 So AH is perpendicular to BC. Similarly for other vertices and their opposite sides. Since H lies on each altitude, it's the orthocenter.