If ABC is a triangle whose orthocentre is P and the circumcentre is Q, then prove that $¯ ¯¯¯¯¯¯ ¯ P A + ¯ ¯¯¯¯¯¯ ¯ P C + ¯ ¯¯¯¯¯¯ ¯ P B = 2 ¯ ¯¯¯¯¯¯ ¯ P Q$ .

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Solution

As, $2 Q G = P G$
( $G$ is centroid)
Let $D$ is pt. on $B . C$
$Q D = \frac{Q B + Q C}{2}$
& $Q G = \frac{Q A + 2 Q D}{3}$
$\Rightarrow 3 Q = Q A + Q B + Q C$
$\Rightarrow 2 P Q = Q A + Q B + Q C$

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Δ

\to P A + \to P B + \to P C = ?

¯ ¯¯¯¯¯¯ ¯ Q A + ¯ ¯¯¯¯¯¯¯ ¯ Q B + ¯ ¯¯¯¯¯¯¯ ¯ Q C = ¯ ¯¯¯¯¯¯ ¯ Q P

Δ A B C

- - \to O A + - - \to O B + - - \to O C = - - \to O P

- - \to P A + - - \to P B + - - \to P C = 2 - - \to P O

∠ A B C = 90^{\circ}

∠ P C B = ∠ P B C

¯ ¯¯¯¯¯¯ ¯ Q P = 3 ¯ ¯¯¯¯¯¯¯ ¯ Q G

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