Dear student To Prove: The lines joining the vertices of a tetrahedron to the centroids of the opposite faces are concurrent Proof:Let a, b, c, and d are the position vector of the vertices of tetrahedron ABCD Let G1,G2,G3, and G4 are the centroid of ΔBCD,ΔCDA,ΔABD and ΔABC respectively Now,Position vector of G1=13b+c+dAnd, position vector of G dividing AG1 in the ratio 3:1 So,OG→=3×OG1→+1×OA→3+1=3×13b+c+d+1×a3+1=14a+b+c+dSimilarly, by symmetry, AG2,AG3, and AG4 are also divided at the point G in the ratio 3:1Hence, the lines AG1,AG2,AG3, and AG4 are concurrent and passes through the point G ehose position vector is 14 Regards

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T o P r o v e : T h e l i n e s j o i n i n g t h e v e r t i c e s o f a t e t r a h e d r o n t o t h e c e n t r o i d s o f t h e o p p o s i t e f a c e s a r e c o n c u r r e n t .

P r o o f : L e t a, b, c, a n d d a r e t h e p o s i t i o n v e c t o r o f t h e v e r t i c e s o f t e t r a h e d r o n A B C D . L e t G_{1}, G_{2}, G_{3}, a n d G_{4} a r e t h e c e n t r o i d o f Δ B C D, Δ C D A, Δ A B D a n d Δ A B C r e s p e c t i v e l y . N o w, P o s i t i o n v e c t o r o f G_{1} = \frac{1}{3} (b + c + d) A n d, p o s i t i o n v e c t o r o f G d i v i d i n g A G_{1} i n t h e r a t i o 3 : 1 . S o, \vec{O G} = \frac{3 \times \vec{O G_{1}} + 1 \times \vec{O A}}{3 + 1} = \frac{3 \times \frac{1}{3} (b + c + d) + 1 \times a}{3 + 1} = \frac{1}{4} (a + b + c + d) S i m i l a r l y, b y s y m m e t r y, A G_{2}, A G_{3}, a n d A G_{4} a r e a l s o d i v i d e d a t t h e p o i n t G i n t h e r a t i o 3 : 1 H e n c e, t h e l i n e s A G_{1}, A G_{2}, A G_{3}, a n d A G_{4} a r e c o n c u r r e n t a n d p a s s e s t h r o u g h t h e p o i n t G e h o s e p o s i t i o n v e c t o r i s \frac{1}{4} .

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