Question

If ABC is a triangle whose orthocenter is P and the circumcenter is Q, then prove that vectors PA + PC + PB = 2 PQ

Answer 1

Let G be the centroid of the ∆ ABC.

Let A, B, C, G, Q have position vectors \(\overline{a},\) \(\overline{b},\) \(\overline{c},\) \(\overline{g},\) \(\overline{q}\) w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.