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

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Solution

$G i v e n t h e o r t h o c e n t r e o f a △ l e = P$
$a n d c i r c u m c e n t r e = Q$
$W e h a v e t o p r o v e Q A + Q B + Q C = Q P$
$w e k n o w t h a t Q G = 2 G P,$
$w h e r e G i s t h e c e n t r o i d o f △ l e .$
$L e t a p o i n t D b / w B a n d C$
$⟹ Q D = \frac{Q B + Q C}{2}$
$∴ Q A + Q B + Q C = Q A + 2 Q D$
$W e k n o w t h a t G d i v i d e s t h e p o i n t A a n d m i d$
$p o i n t o f o p p o s i t e s i d e (D) i n r a t i o 2 : 1.$
$Q G = \frac{Q A .2 Q D}{2 + 1}$
$⟹ Q A + Q B + Q C = 3 Q G$
$= 2 Q G + Q G$
$= G P + Q G = P Q$
$∴ ¯ ¯¯¯¯¯¯ ¯ Q A + ¯ ¯¯¯¯¯¯¯ ¯ Q B + ¯ ¯¯¯¯¯¯¯ ¯ Q C = ¯ ¯¯¯¯¯¯ ¯ Q P$

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¯ ¯¯¯¯¯¯ ¯ P A + ¯ ¯¯¯¯¯¯ ¯ P C + ¯ ¯¯¯¯¯¯ ¯ P B = 2 ¯ ¯¯¯¯¯¯ ¯ P Q

¯ ¯¯¯¯¯¯ ¯ Q P = 3 ¯ ¯¯¯¯¯¯¯ ¯ Q G

Q

A B C

\to Q A + \to Q B + \to Q C = \to 0

A (¯ ¯ ¯ a)

B (¯ ¯ b)

C (¯ ¯ c)

A B C

p + q + r = 0 = a + b + c,

△ = ∣ ∣ ∣ ∣ \begin{matrix} p a & q b & r c q c & r a & p b r b & p c & q a \end{matrix} ∣ ∣ ∣ ∣

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