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Question
If ABC is a triangle whose orthocentre is P and the circumcentre is Q, prove that `bar"PA" + bar"PB" + bar"PC" = 2bar"PQ".`
Solution
Let G be the centroid of the Δ ABC.
Let A, B, C, G, Q have position vectors `bar"a",bar"b",bar"c",bar"g",bar"q"` w.r.t. P. We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2.
∴ `bar"g" = (1. bar"p" + 2bar"q")/(1 + 2) = (2 bar"q")/3 .........[∵ bar"p" = bar0]`
∴ `3bar"g" = 2bar"q"`
∴ `(3(bar"a" + bar"b" + bar"c"))/3 = 2bar"q"`
∴ `bar"a" + bar"b" + bar"c" = 2bar"q"`
∴ `bar"PA" + bar"PB" + bar"PC" = 2bar"PQ"`
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