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The Centroid

If the median...

Question

If the medians of $Δ A B C$ intersects at $G$ , then $a r (Δ A G B) = a r (Δ A G C) = a r (Δ B G C) = \frac{1}{k} a r (Δ A B C)$ .Find k

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Solution

$G i v e n : T h e m e d i a n s A D, B E a n d C F o f △ A B C i n t e r s e c t s a t G W e k n o w t h a t m e d i a n s o f a t r i a n g l e d i v i d e s i t i n t o t w o t r i a n g l e s o f e q u a l a r e a . I n △ A B C, A D i s t h e m e d i a n \Rightarrow a r (△ A B D) = a r (△ A C D) . . . . . . . (1) I n B G C, G D i s t h e m e d i a n \Rightarrow a r (△ G B D) = a r (△ G C D) . . . . . (2) S u b s t r a c t i n g (2) f r o m (1), w e g e t a r (△ A B D) - a r (△ G B D) = a r (△ A C D) - a r (△ G C D) \Rightarrow a r (△ A G B) = a r (△ A G C) . . . . . . (3) S i m i l a r l y, a r (△ A G B) = a r (△ G B C) . . . . . (4) F r o m (3) a n d (4) w e g e t a r (△ A G B) = a r (△ A G C) = a r (B G C) B u t, a r (△ A G B) + a r (△ A G C) + a r (△ B G C) = a r (△ A B C) ∴ 3 a r (△ A G B) = a r (△ A B C) \Rightarrow a r (△ A G B) = \frac{1}{3} a r (△ A B C) H e n c e, a r (A G B) = a r (A G C) = a r (B C G) = \frac{1}{3} a r (A B C)$

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